CALCULATION EXERCICES

To the LABORATORY EXPERIMENTS IN MEDICAL
CHEMISTRY

Edited by:

Zoltán Matus

Compiled by:

Péter Jakus
László Márk
Anikó Takátsy

Pécs, 2007

Table of content:

Introduction

3

1.

Stoichiometry I. Balancing equations

4

2.

Stoichiometry II. Calculation exercises

9

3.

The gaseous state

13

4.

Concentrations of solutions

17

5.

Calculations connected to titrimetry

28

6.

Electrolytic dissociation

33

7.

Dilute solutions

38

8.

Hydrogen ion concentration, pH, buffers

44

9.

Heterogeneous equilibria. Crystallisation, solubility product, partition coefficient

55

10.

Thermochemistry

64

11.

Electrochemistry

67

2

INTRODUCTION
The chapter is devoted to helping the students practice the most important topics of
General Chemistry. The order of the sections follows the schedule of the lectures and seminars, and their volume indicates the importance of the topic. Each section begins with a few solved problems. They represent the minimum requirement at the exam. The worked-out solutions are not the only ones. For an easier self-checking, the numerical results of the unsolved calculation exercises are given in parentheses after each question.

Sources:

1.) Laboratory experiments in medical chemistry, ed. György Oszbach, Pécs, 1998.
2.) Villányi Attila: Ötösöm lesz kémiából, (6. ed.) Mőszaki Könyvkiadó, Budapest., 1999
3.) Charles E. Mortimer, Ulrich Müller: Chemie- Das Basiswissen der Chemie (8.komplett überarbeitete und erweiterte Auflage), Thieme, 2003
4.) Villányi Attila: Kémia feladatgyőjtemény a kétszintő érettségire. Kemavill. Bt., Budapest,
2004
5.) Web page of Technical University of Budapest for 1st year students: http://web.inc.bme.hu/fpf/index.html 3

1. STOICHIOMETRY I
EXERCISES ON WRITING EQUATIONS
Complete the following skeletal equations with coefficients. The ones indicated by an asterisk should be transformed into ionic equations.
1.

N2 + H2 = NH3

2.

Al2O3 + HCl = AlCl3 + H2O

3.

SiCl4 + H2O = SiO2 + HCl

4.

CaCO3 + HCl = CaCl2 + H2O + CO2

5.

CS2 + NaOH = Na2CS3 + Na2CO3 + H2O

6.

K + H2O = KOH + H2

7.*

aqueous sodium hydroxide

+

hydrochloric acid solution

=

aqueous sodium

chloride + water
8.*

aqueous ammonium sulphate + aqueous potassium hydroxide = ammonia gas + aqueous potassium sulphate + water

9.*

aluminium metal + hydrochloric acid solution = aqueous aluminium chloride + hydrogen gas

10.* aqueous silver sulphate + aqueous barium iodide = solid silver iodide +
+ solid barium sulphate.
11.* aqueous magnesium chloride + aqueous sodium hydroxide = solid magnesium hydroxide + aqueous sodium chloride
12.* aqueous iron(II) sulphate + aqueous ammonium sulphide = solid iron(II) sulphide
+ aqueous ammonium sulphate
13.* aqueous iron(III) chloride + aqueous sodium hydroxide = solid iron(III) hydroxide
+ aqueous sodium chloride
14.* solid calcium carbonate + hydrochloric acid = aqueous calcium chloride + carbon dioxide gas

Redox reactions
I.

Determine the oxidation state of the S and C atoms in the following compounds: sodium tetrathionate, butadiene, ethanol and acetic acid.

II.

Determine the stoichiometric coefficients of the following redox reactions and write balanced ionic equations.

1.

H2O2 + HI = I2 + H2O

2.

I2 + Na2S2O3 = NaI + Na2S4O6

3.

NaOCl = NaClO3 + NaCl

4.

Br2 + NaOH = NaBr + NaOBr + H2O

5.

HNO2 = HNO3 + NO + H2O

6.

KMnO4 + H2O2 + H2SO4 = MnSO4 + K2SO4 + H2O + O2

7.

KMnO4 + FeSO4 + H2SO4 = MnSO4 + K2SO4 + Fe2(SO4)3 + H2O

8.

KMnO4 + H2SO3= MnSO4 + K2SO4 + H2SO4 + H2O

9.

K2Cr2O7 + KI + H2SO4 = Cr2(SO4)3 + I2 + K2SO4 + H2O

10.

FeCl3 + KI = I2 + FeCl2 + KCl

11.

KOH + Cr2O3 + KNO3 = K2CrO4 + KNO2 + H2O

12.

K2Cr2O7 + FeSO4 + H2SO4 = Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + H2O

13.

(COOH)2 + KMnO4 + H2SO4 = CO2 + MnSO4 + K2SO4 + H2O

14.

HCOOH + KMnO4 = MnO2 + CO2 + H2O + KOH

15.

MnO2 + HBr = MnBr2 + Br2 + H2O

16.

MnO2 + KNO3 + KOH = K2MnO4 + KNO2 + H2O

17.

SnCl2 + KBrO3 + HCl = SnCl4 + KBr + H2O

18.

NH3 + O2 = N2 + H2O

19.

NH3 + O2 = NO + H2O

20.

Fe + O2+ H2O = Fe(OH)3

21.

I2 + SO32- = I- + SO42-

22.

IO3- + I- + H+ = I2 + H2O

23.

IO3- + SO32- + H2O = I- + SO42- + H+

24.

ClO3- + I- + H+ = Cl- + I2 + H2O

25.

NO2 + H2O = H+ + NO3- + NO2-

26.

I2 + S2- + H2O = SO42- + I- + H+

27.

Br2 + SO2 + H2O = Br- + SO42- + H+

28.

NO3- + Cu + H+ = Cu2+ + NO + H2O

5

29.

MnO2 + Cl- + H+ = Mn2+ + Cl2 + H2O

30.

MnO4- + Cl- + H+ = Mn2+ + Cl2 + H2O

31.

MnO4- + SO2 + H2O = SO42- + Mn2+ + H+

32.

MnO4- +

33.

MnO4- + S2- + H2O = MnO2 + S + OH-

34.

MnO4- + NO2- + H+ = Mn2+ + NO3- + H2O

35.

Cr2O72- + H2S + H+= Cr3+ + S + H2O

36.

CrO42- + I- + H+ = Cr3+ + I2 + H2O

37.

Cr2O72- + H2O2 + H+= Cr3+ + O2 + H2O

38.

Cr2O72- + CH3OH + H+ = Cr3+ + CO2 + H2O

39.

Pb(NO3)2 = PbO + NO2 + O2

40.

Pb + PbO2 + H2SO4 = PbSO4 + H2O

41.

S + NaOH = Na2S + Na2S2O3 + H2O

42.

Ag + HNO3 = AgNO3 + NO + H2O

43.

Cu + H2SO4= CuSO4 + SO2 + H2O

44.

As2S3 + NH3 + H2O2 = (NH4)3AsO4 + S + H2O

45.

KBrO3 + H3AsO3 = KBr + H3AsO4

I- + H2O = IO3- + MnO2 + OH-

Acid-base reactions
Write out as many acid-base reactions as you can combining any reactants according to the scheme below.
H+ + A- + B+ + OH-

The stoichiometric equation:

HA + BOH

The ions existing in the solution:

H+ + A- + B+ + OH- = H2O + A- + B+

The ionic equation:

H+ + OH- = H2O

Reactions with the evolution of a gas
Complete and balance the following skeletal equations.
1.

FeS + H+ = H2S + Fe2+

2.

Fe + H+ = H2 + Fe2+

3.

Zn + H+ = H2 + Zn2+

6

4.

SO32- + H+ = H2O + SO2

5.

CaCO3 + H+ = Ca2+ + H2O + CO2

6.

HCO3- + H+ = H2O + CO2

7.

NH4+ + OH- = NH3 + H2O

8.

CO32- + H+ = CO2 + H2O

9.

S2- + H+ = H2S

Reactions with a precipitate formation
Complete and balance the following skeletal equations.
1.

Ca2+ + PO43- = Ca3(PO4)2

2.

Al3+ + OH- = Al(OH)3

3.

Pb2+ + Cl- = PbCl2

4.

Ag+ + SO42- = Ag2SO4

5.

Ag+ + CrO42- = Ag2CrO4

6.

Ag+ + S2- = Ag2S

7.

As5+ + S2- = As2S5

8.

Ba2+ + SO32- = BaSO3

9.

Ca2+ + F- = CaF2

10.

Cu+ + S2- = Cu2S

11.

Mg2+ + CO32- = MgCO3

12.

Ni2+ + S2- = NiS

13.

Zn2+ + OH- = Zn(OH)2

Dissolution of a precipitate in an acid-base reaction
Complete and balance the following skeletal equations.
(NO3-) = Hg22+ + H2O

1.

Hg2O + H+

2.

HgO + H+

3.

Fe(OH)2 + H+ =

4.

Fe(OH)3 + H+ = Fe3+ + H2O

5.

As2S3 + S2-

6.

FeS + H+

(NO3-) = Hg2+ + H2O

=

Fe2+ + H2O
AsS33-

= Fe2+ + H2S

7

7.

CaCO3 + H+

= Ca2+ + CO2 + H2O

8.

Mg(OH)2 + H+

9.

BiOCl + H+

10.

Zn(OH)2 + OH-

[Zn(OH)4]2-

11.

Pb(OH)2 + OH-

[Pb(OH)4]2-

12.

Al(OH)3 + OH-

[Al(OH)4]-

= Mg2+ + H2O
Bi3+ + H2O + Cl-

Dissolution of a precipitate by complex formation
Complete and balance the following skeletal equations.
1.

HgI2 + I-

[HgI4]2-

2.

Fe(CN)2 + CN-

3.

Fe(CN)3 + CN-

4.

AgCl + NH3

=

[Ag(NH3)2]+

5.

AgI + S2O32-

=

[Ag(S2O3)2]3- + I-

6.

AgCl + S2O32-

=

[Ag(S2O3)2]3- + Cl-

7.

AgBr + S2O32-

=

[Ag(S2O3)2]3- + Br-

=
=

[Fe(CN)6]4[Fe(CN)6]3+ Cl-

8

2. STOICHIOMETRY II

CALCULATION EXERCISES

1.)

How many grams are 5 moles of elementary sodium?
The molar mass of sodium is: 23 g/mol
1 mole sodium

23 g

5 mole sodium

xg x = 115

Thus, 5 mol sodium is 115 g.
2.)

How many grams and moles of CuO are formed by the oxidation of 100 g Cu?

a.)

The reaction is the following: 2 Cu + O2 = 2 CuO

According to the equation:

1 mol CuO is formed from 1 mol Cu, i.e.

63.5 g Cu is oxidized to

79.5 g CuO

100 g Cu is oxidized to

x g CuO

63.5:100 = 79.5:x x =

7950
= 125 g
63.5

Thus, 100 g Cu can be oxidized to form 125 g CuO.
b.)

79.5 g CuO is

1 mol

125 g CuO is

x mol
79.5:125 = 1:x x= 125
= 1.6
79.5

Thus, 125 g CuO is 1.6 mol.

9

3.)

Calculate the percentage of water in crystalline calcium chloride.
The formula:

CaCl2 · 6 H2O

The formula mass: 40 + 71 + (6 · 18) = 219 g/mol
According to the formula, 1 mol (219 g) crystalline calcium chloride contains
6 moles (108 g) of water.
219:108 = 100:x x =

10800
= 49
219

Thus, the water content of crystalline calcium chloride is 49 m/m %.
4.)

Calculate the molar mass of the following compounds: H2O, NO2, CO2, CS2, C6H6. (18;
46; 44; 76; 78 g/mol)

5.)

How many atoms are there in
a) 1 molecule of glucose,

(24)

b) 1 mol formic acid,

(3 · 1024)

c) 32 g oxygen,

(1.2 · 1024)

6.)

How many moles are there in 1.9 kg iodine?

7.)

How

many

moles

are

there

in

1000

(7.5) g crystalline

(Na2S2O3 · 5 H2O). How many moles of water does it contain?

sodium

thiosulphate

(4; 20)

(3.3 · 1019)

8.)

How many molecules are there in 1 mg water?

9.)

How many grams of fluorine are there in 10 g cryolite (Na3AlF6)?

10.) What is the percentage of water in borax (Na2B4O7 · 10 H2O)?

(5.43)
(52.7)

11.) How many grams of crystalline copper(II) sulphate (CuSO4 · 5 H2O) is required to be weighed to obtain 300 g anhydrous copper(II) sulphate?

(471)

12.) How many moles of electrons are released by 70 g iron(II) ion during oxidation, and how many iron(III) ions form?

(1.25; 7.52 · 1023)

13.) How many g NaOH forms placing 4.6 g sodium into water?

(8)

14.) How many g lead forms if 32.3 g PbO is reduced with hydrogen?

(30)

15.) How many moles of sulfuric acid are needed to dissolve 1.5 g aluminium? (0.083)
16.) How many g table salt is needed to produce 43.7 g HCl?

(70.05)

10

17.) How many g of water is necessary to dilute 200.0 g 50.0 m/m% sulfuric acid solution to
10.0 m/m%?

(800)

18.) What is the percentage mass loss of potassium chlorate on heating?
(2 KClO3 = 2 KCl + 3 O2)

(39.2)

19.) From 100.0 g 35.0 m/m% sugar solution some water was evaporated and the mass of the solution decreased by 6.0 g. What is the m/m% composition of the residue?
(37.2)
20.) What is the percentage of water in a tissue if a 1.20-g sample, when heating to a constant weight, gives a residue of 308 mg?

(74.33)

21.) What is the percentage of sulphur in a mixture of 10.0 g anhydrous sodium sulphate and of an equal amount of crystalline sodium sulphate (Na2SO4 · 10 H2O)?
22.) What amount of phosphoric acid (H3PO4) contains 10 g of phosphorous?

(16.16)
(31.6)

23.) An unknown compound has a molar mass of 46 g/mol. What is the formula of it if the elemental analysis results are: C: 52.2 %, H: 13.0 %, O: 34.8 %? (C2H6O)
24.) What is the formula of the substance of the elementary composition C: 65 %, H: 13.5
%, O: 21.5 %?

(C4H10O)

25.) By the combustion of 2.50 g of nicotine 6.78 g CO2, 1.94 g H2O and 0.432 g of N2 form. What is the m/m% composition of nicotine?

(74% C; 8,6% H; 17,3% N)

26). A sulphur compound of copper contains 20.12% S. Determine the formula of the compound and the oxidation states of the atoms.

(Cu2S; +1; -2)

27.) How many moles of hydrogen gas form if 4.00 moles of iron react with 5.00 moles of water? (Balance the reaction equation.)

(5)

Fe + H2O = Fe3O4 + H2
28.) 1 g Al powder is allowed to react with 16 g iodine. How many grams of aluminium iodide form? Which is the limiting reactant and what is the percentage excess of the other? (15.11; 13.4)

29.) The quantitative determination of CO is based on the following reaction:
I2O5 + CO = I2 + CO2
Balance the equation and calculate the mass of the starting CO if the product is 0.192 g
I2.

(0.106)

11

30.) The saturated solution of ammonium sulfate is 43.0 m/m% at 20°C. Does the whole amount of 200.0 g of ammonium sulfate dissolve placing it into 350.0 g 10 m/m% solution? What is the percentage of the formed solution?

(yes; 42.7%)

31.) For the treatment of a broken leg, 1.45 kg plaster of Paris (CaSO4 · 1/2 H2O) has been used. During solidification, crystalline gypsum (CaSO4 · 2 H2O) forms. Calculate the weight of the cast.

(1.72)

32.) How many g of water forms during the combustion of 1 kg of propane-butane gas mixture if the propane-butane ration is 4:3?

(1594.4)

33.) The hemoglobin contains 0.342% iron. What is the molar mass of hemoglobin if one molecule contains 4 irom atoms?

(6.53 · 104)

34.) How many cm3 of 98.0% m/m sulfuric acid solution with a density 1.83 g/cm3 and how many cm3 of water are needed to have 2.0 dm3 20 m/m%, 1.14 g/cm3 solution?
(254.3; 1814)

12

3. THE GASEOUS STATE

1.)

Calculate the density of N2 gas at 0 °C and 101.325 kPa pressure. d= m (mass)
V (vol ume)

It is known that the molar volume of an ideal gas at STP is 22.41 dm3.
22.41 dm3 (1 mole) of nitrogen weighs 28 g.

Thus, its density is:
2.)

d=

28
= 1.25 g/d m 3
22.41

Calculate the volume of hydrogen gas produced at 27 °C and 98.66 kPa if 1 g sodium reacts with ethanol.
The equation:

2 Na + 2 C2H5OH = 2 C2H5ONa + H2

2 mol sodium produces

1 mol H2, i.e.

46 g sodium produces

22.41 dm3 hydrogen at STP

1g

"

x

dm3 hydrogen at STP

46 : 1 = 22.41 : x

x=

22.41
= 0.487
46

Thus, at 0 °C and 101.325 kPa pressure, 0.487 dm3 hydrogen evolves.
Applying the general gas law, we may calculate the volume at 27 °C and
98.66 kPa. po = 101.325 kPa

p1 = 98.66 kPa

To = 273 K

T1 = 300 K

vo = 0.487 dm3

v1 = ?

v1 =

101, 325 ⋅ 0 , 487 ⋅ 300
= 0 , 55
273 ⋅ 98 , 66

Thus, 1 g Na at 27 °C and 98.66 kPa produces 0.55 dm3 H2 gas.

13

3.)

A gas mixture at 101.325 kPa pressure is composed of 35 v/v % oxygen and 65 v/v % nitrogen. Calculate the partial pressures of the components.
According to Dalton's law, the total pressure equals the sum of the partial pressures:

p mi x t u r e = p O 2 + p N 2
The partial pressures are proportional with the volumes of the components:

pO2 pN2 Thus,

=

VO 2
VN 2

pO2 = p N 2

=

35
65

35
65

p O 2 value substituted into Dalton's law:
101.325 = p O 2 + p N 2 =

35 pN + pN2
65 2

6586 = ( 35 + 65) p N 2 pN2 =

6586
= 65.86
100

Thus, the partial pressure of nitrogen is 65.86 kPa, while that of oxygen is 35.46 kPa.
4.)

What are the densities of the following gases among normal conditions: C3H8, CH4, HI?
(1,96; 0,71; 5,71)

5.)

How many molecules are there in 1 cm3 H2 gas at STP?

(2.68 · 1019)

6.)

Calculate the mass of 1 cm3 of dinitrogen oxide gas at 506.63 kPa and 20 °C.
(9.15 mg)

7.)

Calculate the molar mass of a gas with a density of 0.5 related to oxygen.

(16)

8.)

The volume of a gas is 500 cm3 at 100°C and 100.26 kPa pressure. What is the volume of this gas among normal conditions? (362.1)

9.)

2.75 g of an organic compound is evaporated at 150 °C and 8.28 kPa pressure to form
11.66 dm3 vapour. Elemental analysis: C: 24.3 %, H: 4.04 %, Cl: 71.7 %. Calculate the molecular formula of the compound. (C2H4Cl2)

14

10.) Calculate the molecular formula of the compound composed of 84.2 m/m % sulphur and 15.8 m/m % carbon, and its relative density related to nitrogen is 2.71.

(CS2)

11.) A gas mixture contains 2 g of hydrogen and 10 g of methane. The total pressure of the mixture is 253.31 kPa. Calculate the partial pressure of methane.

(60.31)

12.) In a container of compressed air, the pressure is 2026.5 kPa. Calculate the partial pressures of O2 and N2, if air is regarded as a mixture of these two gases.
(405.3; 1621.2)
13.) 1 dm3 methane is combusted at STP. How many grams of CO2 and H2O form?
(1.96; 1.61)
14.) How many dm3 ammonia evolve from 0.03 kg ammonium chloride reacted with sodium hydroxide at 20 °C and 99.99 kPa?

(13.7)

15.) A container of 30 L contains oxygen gas at 20 °C and 313.97 kPa pressure. Some part of the gas is disinflated. When the residue of the gas reaches again the 20 °C its pressure is 243.18 kPa. How many g of oxygen was disinflated from the container? (1945.28)
16.) The mass of of 1 L of carbon dioxide is 1.9766 g among normal conditions. Calculate the pressure where the mass of 1 L of carbon dioxide is 1 g if the temperature has a constant value. (51.26)
17.) In a space capsule, the astronauts expire 8750 dm3 air of 4.1 % v/v CO2 content daily at
23 °C and 70 kPa. The expired air is reacted with LiOH. How many days can they spend in the space started with 80 kg LiOH? (163)
18.) Arsine (AsH3) decomposes thermally into its elements. How many dm3 arsine gas of
18 °C and 99.99 kPa pressure should be decomposed thermally to produce 10 dm3 hydrogen gas of 40 °C and 101.325 kPa ?

(6.28)

14.) How many dm3 0.20 molar KMnO4 solution is required for the removal of the H2S content of 15.0 dm3 gas mixture at 33 °C, if the partial pressure of H2S is 50 kPa? An incomplete skeletal equation is given below:
KMnO4 + H2S = KOH + K2SO4 + MnO2

(4)

15.) To 200 g 18.25 m/m % hydrochloric acid, 25 g calcium carbonate is added. Calculate the volume of evolved carbon dioxide at 20 °C and 0.1 MPa and the final percentage of hydrochloric acid solution.

(6; 8.52)

15

4. CONCENTRATION OF SOLUTIONS

1.)

50 cm3 KOH solution is made of 14 g KOH. The density of the solution at 20 °C is 1.22 g/cm3. Calculate the m/v %, m/m % and g/dm3 concentrations, the molarity and molality of the solution.
Calculation of percentage:
50 cm3 solution contains

14 g KOH

100 cm3 solution contains

x g KOH

50:100 = 14:x x = 28

Thus, the percentage of the KOH solution is 28 m/v%.
Calculation of m/m %:

d = 1.22 g/cm3

the mass of 50 cm3 solution = 50.1.22 = 61 g
61 g solution contains

14 g KOH

100 g solution contains

x g KOH

61:100 = 14: x x= 1400
= 23
61

Thus, the mass per cent concentration of the solution is 23 m/m %.
Calculation of g/dm3 concentration:
50 cm3 solution contains

14 g KOH

1000 cm3 solution contains

x g KOH

50:1000 = 14 x x = 280

Thus, the KOH solution is of 280 g/dm3 concentration.
Calculation of molarity:
50 cm3 solution contains

14 g KOH

1000 cm3 solution contains

x g KOH

x = 280 (see the g/dm3 value)

16

56 g KOH

1 mol

280 g KOH

x mol

56 : 280 = 1 : x

280
=5
56

x=

Thus, the KOH solution is of 5 M.
Calculation of molality:

d = 1.22 g/cm3

the mass of 50 cm3 solution is 61 g
61 g solution contains 14 g solute and 61-14 = 37 g solvent.
47 g solvent dissolves

14 g KOH

1000 g solvent dissolves

x g KOH

47:1000 = 14: x x= 14000
= 298
47

Thus, 1000 g solvent dissolves 298 g KOH.
56 g

1 mol

298 g

x mol

56 : 298 = 1 : x x= 298
= 5.32
56

Thus, the molality of the solution is 5.32 mol/kg solvent.
2.)

How many cm3 2 % sodium hydroxide solution should be mixed with 0.5 dm3 6 % sodium hydroxide solution to obtain a 3 % solution?
In the laboratory work it is a frequent task to prepare solutions by the dilution of the available concentrated solutions or by mixing solutions of different concentrations in a calculated proportion. When applying the law of conservation of mass for solutions, the following equations can be used:

17

Equation of mixing: c1m1 + c2m2 + c3m3 +...= como m1 + m2 + m3 +...= mo where: c = concentration

m = mass of the solution

Equation of dilution (c2 = 0): c1m1 = como
In the case of not too concentrated aqueous solutions d = 1 g/cm3. Then m ≈ v, and mass can be replaced by volume.

The solution of the above problem: c1 = 2 %

v1 = x

c2 = 6 %

v2 = 500 cm3

co = 3 %

vo = v2 - x

4x + 6 · 500 = 3 (500 + x) x = 3000 - 1500 = 1500

Thus, 500 cm3 6 % sodium hydroxide solution should be mixed with 1500 cm3 2 % sodium hydroxide solution to obtain a 3 % solution.
3.)

How much distilled water and 30 % hydrogen peroxide solution are needed to prepare 6 dm3 1.5 % hydrogen peroxide solution? c1 = 30 %

v1 = x

co = 1.5 %

vo = 6000 cm3
30x = 6000 · 1.5 x = 300

Thus, 6 dm3 1.5 % hydrogen peroxide solution can be made from 300 cm3 concentrated solution and 5700 cm3 distilled water.
4.)

Calculate the mass per cent of a solution made by the dissolution of 5 g sugar in 73 g water. (6.41)

18

5.)

Calculate the volume per cent of an aqueous acetone solution if 125 cm3 solution contains 1.4 cm3 acetone.

6.)

(1.12)

Calculate the percentage of urea in a urine sample, 75 cm3 of which contains 180 mg dissolved urea (carbamide).

(0.24)

7.)

How could you prepare 250 cm3 0.02 M NaOH solution?

(from 0.2 g NaOH)

8.)

Calculate the molarity of a 12.5 % sulphuric acid solution.

(1.275)

9.)

Calculate the g/dm3 concentration of a 0.07 M CaCl2 solution.

(7.76)

10.) 56 g KOH is dissolved in 500 g water. Calculate the m/m % of the solution.
(10.07)
11.) Calculate the m/m% of a solution made by the addition of 20 g solute to 180 g
10 m/m% solution.

(19)

12.) How many grams of solute are dissolved in 500 cm3 of each a.) 0.25 M FeSO4 and b.)
10-2 M CaCl2 solutions?

(18.99; 0.555)

13.) 12 g KOH is dissolved in 500 g water. Calculate the m/m%, the molality and the molar
(2.34; 0.429; 7.65 · 10-3)

fraction of the solution.

14.) Calculate the m/m% concentration of the 1 molal sulphuric acid solution.
(8.925)
15.) What volume of an 8 v/v% solution contains the same amount of solute as 200 cm3 of a
12 v/v% solution?

(300)

16.) From 200 g 8 m/m % sulphuric acid solution 40 g water is evaporated. The density of the remaining solution is 1.400 g/cm3. Calculate the m/m% and the mol/dm3 concentrations of the remainder.

(10; 1.43)

17.) How could you make 150 cm3 0.12 M NaOH solution? What is the percentage of this solution? (0.72; 0.48)

18.) How many cm3 of sulfuric acid of 1.830 g/cm3 density and 98 m/m% is needed to prepare 750 cm3 0.5 M H2SO4 solution?

(20.49)

19.) Calculate the % and the molar concentrations of the Ca(OH)2 solution, 22.5 cm3 of which contains 1.3 g Ca(OH)2? (5.78; 0.78;)

19

20.) Calculate the molarity and the molality of the 5.1 m/m % H2SO4 solution.
(d = 1.032 g/cm3) (0,537; 0,548; 1,86)
21.) Calculate the molarity and molality of a 15.95 m/m % table-sugar solution of
1.063 g/cm3 density (C12H22O11).

(0.5; 0.55)

22.) The 2.46 mol/dm3 ammonium sulphate solution is of 28.0 m/m %. Calculate the density and the molality of the solution.

(1.160; 2.94)

23.) Calculate the molarity, molality and m/m% of a phosphoric acid solution with a molar fraction 0.216 and 1.426 g/cm3 density?

(8.73; 15.31; 60)

24.) Calculate the molarity and m/m% of a perchloric acid solution of 2.0 molality and 1.100 g/cm3 density. (1.83; 16.73)
25.) Calculate the molar mass of an acid if its 17.59 m/m% solution is of 1.10 g/cm3 density and 3.07 mol/dm3.

(63; 5.75)

26.) 50 kg KNO3 should be diluted to 10 m/m% by the addition of an indifferent solid. How much solid is to be added?

(450)

27.) How much water should be evaporated from 10 kg 30 m/m% NaOH solution to obtain a
50 m/m% solution?

(4)

28.) By what factor should a 160 g/dm3 acetic acid solution be diluted to prepare an 0.10 M solution? (26.7)

29.) Calculate the percentage of the HNO3 solution made by the 15-fold dilution of a 12 %
HNO3 solution.

(0.8)

30.) Calculate the percentage of the solution made by the addition of 30 cm3 water to
135 cm3 2.5 % Na2S2O3 solution.

(2.05)

31.) How many grams of 10 m/m% and 96 m/m% sulphuric acid solutions are to be mixed for the preparation of 200 g 30 m/m% solution?

(153.5; 46.5)

32.) How many grams of 4 m/m% CuSO4 are to be used for the dissolution of 200 g
CuSO4 · 5 H2O to form a 16 m/m% solution? (799)
33.) How many cm3 34.4 m/m% hydrochloric acid solution of 1.175 g/cm3 density are required to make 5 dm3 0.1 molar solution?

(45.15)

34.) How many cm3 (d = 1.830 g/cm3) of 93.6 m/m% sulphuric acid are required to prepare
1 dm3 9.80 m/m% (d = 1.065 g/cm3) solution by dilution with water?

(60.9)
20

35.) 100 cm3 K2CO3 solution of 24.0 m/m% concentration and 1.232 g/cm3 density is diluted with 50.0 g water. Calculate the molality of the diluted solution.

(1.49)

36.) 500 cm3 2.0 molar HCl solution of 1.034 g/cm3 density should be prepared. How many cm3 36.0 m/m% hydrochloric acid solution of 1.180 g/cm3 density and how many cm3 water are required?

(85.9; 416)

37.) How many cm3 30 m/m% sodium hydroxide solution (d = 1.33 g/cm3) are required to produce 750 cm3 2 mol/dm3 solution? (150.38)
38.) To 3 dm3 5 m/m% H2SO4 solution (d = 1.032 g/cm3), 2 dm3 50.5 m/m% H2SO4 solution
(d = 1.39 g/cm3) is added. Calculate the density of the resulting solution if its concentration is 3.2 M. (1.18)
39.) 250 cm3 5.00 mol/dm3 potassium hydroxide solution is mixed with 150 cm3
25.40 m/m% sulphuric acid solution of 1.185 g/cm3 density. Then, the volume is made up to 1.0 dm3. Calculate the mol/dm3 concentrations of the reaction product and of the excess reagent.

(0.461; 0.328)

40.) Lactic acid is miscible with water. Calculate the mass per cent composition of the lactic acid solution in which the mol % value is half of the m/m % value. How many grams of
NaOH can neutralize as many grams of the lactic acid solution as its mass per cent value is? (75.0; 25.0)

41.) In 150 g 10 m/m % NaOH solution 22 g sodium is dissolved. Calculate the mass per cent of the resulting NaOH solution.

(31.33)

42.) How many grams of sulphur trioxide have been dissolved in 60 g 4.9 mass per cent sulphuric acid solution if the final solution is of 12.25 m/m %?

(4.0)

43.) 200 g 10 m/m% sulphuric acid solution is mixed with 200 g 10 m/m% sodium hydroxide solution. How many grams and of which reactant remain unchanged after the completion of the reaction? Is the resulting solution basic or acidic?
(3.67 g NaOH; basic)
*44.) 318.0 g sodium carbonate solution is neutralized with 49.0 m/m% sulphuric acid solution. The sodium sulphate concentration of the resulted solution is 14.0 m/m%.
Calculate the m/m% of the sodium carbonate solution.

(12.35)

45.) As a result of administration, for example, penicillin to a penicillin-sensitive patient, a so-called anaphylactic shock may develop. Because of the allergic reaction, the blood

21

vessels dilate to such a degree that the blood virtually gets lost in their volume, so that the supply of the different organs becomes insufficient. In order to avoid imminent death, we have to make the vessels constrict by injecting 1 mg adrenaline intravenously.
The molecular formula of adrenaline is C8H13O3N. In the commercially available ampoule you have a 5.464 mM adrenaline solution. How many cm3 must be injected?
(1.07)
46.) An over-1.74-mole-per-day alcohol intake for a decade or more may cause a liver

cirrhosis. How many litres per day of beer (3.5 m/v%), wine (11.5 m/v%) or of a hard drink (40 m/v% alcohol content) consumption is the upper limit to avoid this serious damage? (2.29; 0.7; 0.2)

47.) As a parenteral nutriment of a child, 500 cm3 10 m/v% sugar solution is required. In the pharmacy, 5 % Isodex and 40 % glucose solutions are available in 500 cm3 sterilized bottles. Propose a method for the preparation of the above solution considering that, to maintain the sterility of the stock solutions, only sterile syringes can be used for transferring. (71.43 cm3 5 % solution sucked out and replaced by 40 % solution.)

48.) A 5-years-old child of 20 kg body weight suffers from cerebral oedema. For decreasing the pressure of the intracranial liquid, 1 g/kg glycerol is administered in every 6 hours.
How many cm3 of 50 % aqueous glycerol should be injected ?

(40)

50.) A sugar solution is prepared from 100 g of water and 40 g of table sugar (C12H22O11).
Calculate the
a.

mass %

b.

mass fraction

c.

mole fraction

d.

molality

e.

molarity of the solution if its density is 1.12 g/cm3.

( 28,57 m/m%; 0,2858; 0,0206; CR=1.17; 0,935M )
51.) How can we prepare 500 mL phosphoric acid solution of 20 m/m% and 1.11 g/cm3 density if we have concentrated phosphoric acid of 60 m/m% and 1.43 g/cm3 density?
(129.37 cm3)
52.) 1 kg of a hydrochloric acid solution in water with a concentration of 2 mol/kg is mixed with 1 kg of 1 mol/kg hydrochloric acid solution. Calculate the Raoult’s concentration of the resulting solution.

(1.491)

22

53.) What is the m/m% of a saturated KNO3 solution at 40°C? At 40 °C 100 g of water dissolves 64 g of KNO3.

(39)

54.) 500 g of saturated KNO3 solution is cooled from 60°C to 20°C. How many grams of
KNO3 were precipitated from the solution? 100 g of water dissolves 110 g of KNO3 at
60 ° and 32 g of KNO3 at 20 °C.

(189.1)

55.) An aqueous solution contains 40 m/m% Na2CO3. Calculate the mol fraction of the salt.
(10.16)
56.) Calculate the Raoult’s concentration of an aqueous solution containing 3 mol% of solute. (1.718)

57.) A naphthalene (C10H8) solution in benzene (C6H6) contains 25 mol% naphthalene.
Calculate the m/m% of the solution.

(35.35)

58.) The Raoult’s concentration of a HCl solution is 10 mol/kg, its density is 1.136 g/cm3.
Give the concentration of the HCl in the following units:
a.

m/m%

b.

mole fraction

c.

mole%

d.

mol/dm3

( 26,72 m/m%; 0,1525; 15,25 mol%; 8,324 M )
59.) A gas mixture is composed by 16 m/m% of oxygen and 84 m/m% of nitrogen. Calculate the composition of the mixture in mol% and v/v%.

(14.3% O2)

60.) The density of a 2 M sulfuric acid solution is 1.120 g/cm3. Calculate the m/m%, mol% and the Raoult’s concentration of the sulfuric acid.

( 17.51 m/m%; 3.75 mol%; CR =

2.164 mol/kg)
61.) The Raoult’s concentration of an aqueous nitric acid solution is 2.5 mol/kg. Calculate the m/m% and the mol% of the nitric acid.

( 13.61 m/m%; 4.308 mol% )

62.) The mole fraction of KCl is 0.24 in an aqueous solution. The density of the solution is
1.12 g/mL. Calculate the Raoult’s concentration, m/m%, m/v% and the molarity of the solution. (CR = 17.6; 56.8 m/m%; 63.2 m/v%; 8,46 M )

63.) Give the concentration of the concentrated, 98% sulfuric acid solution (density: 1.836 g/mL) in mol/L, g/L and mol% units. ( 18.3 M; 1800 g/l ; 90.5 mol% )

23

64.) Calculate the Raoult’s concentration, m/m% and the molarity of a CaCl2 solution if its mole fraction is 0.67, density is 1.324 g/mL. (CR=112.8; 92.53 m/m%; 11.05 mol/dm3)
65.) 100 g NaCl is dissolved in 500 g water. The density of the solution is 1.121 g/mL.
Calculate the m/m%, mol%, the molarity and the Raoult’s concentration of the solution.
( 16,6 % ; 3.196 M ; 5.802 mol% ; CR = 3.422 )
66.) How much water is needed to dissolve 100 g potassium iodide (KI) to prepare a solution of a.

15 %

b.

1.5 mol/kg molality

c.

1.5 mol/dm3 molarity

d.

5 mol%?

(a. 566.66; b. 401.53 c. 368.21 d. 205.2)
67.) How much potassium hydroxide is necessary to prepare 200 mL 0.6 M solution?
(6.732)
68.) A 24 mol% solution of naphthalene (C10H8) in benzene (C6H6) is required. How much naphthalene is needed to prepare 450 g of the solution? Calculate also the number of moles of the solution. (153.6 g C10 H8; 296.6 g C6H6; 5.0 mol)
69.) 100 g sodium chloride is dissolved in such amount of water to form 500 mL of solution.
The density of the solution is 1.132 g/mL. Calculate the molarity, mol% and the
Raoult’s concentration of the solution. (3.422 M; 6.2 mol%; CR=3.671)
70.) 10 mol% aqueous urea (CO(NH2)2) solution is to prepared. How much urea and water are needed for 5 mol of solution? Calculate the Raoult’S concentration of the solution.
(30.03 g urea, 81 g water, CR=6.171)
71.) What amount of sodium hydroxide should be dissolved in water to prepare a
a.

25 m/m%

b.

0.5 mol/dm3 (d = 1.06.g/mL)

c.

2 mol/kg molality solution?

(100; 5.77; 24)

72.) 5 L sulfuric acid of 65.2 m/m% (d = 1.56 g/mL) should be diluted with water to 17 m/m% (d = 1.12 g/mL). How much water is needed? (22.11)
73.) 100 g sulfuric acid of 10 mol% and 150 g sulfuric acid of 15 mol% are mixed. Calculate the mol% composition of the resulting solution.

(12.82)

24

74.) 75 mL of 10 m/m% (d = 1.09 g/mol) and 220 mL of 33 m/m% (d = 1.32 g/mL) potassium hydroxide solutions are mixed. Give the mole fraction of the KOH in the resulting solution.

(0.118)

75.) 30 g of 66 m/m% solution is required. Two solutions, a 78% and a 48% are avaible.
How many g of each solutions should be used?

(18 g of 78 %; 12 g of 48 %)

76.) 100 g of a 15 m/m% sodium hydroxide solution is required. Two solutions, a 585 g/L (d
= 1.44 g/mL) and a 99 g/L (d = 1.100 g/mL) are avaible. How many mL of each solution should be used?

(13.2 mL; 73.6 mL )

77.) 2.5 L KOH solution of 0.4 M is required. A 78 g/L KOH solution (d = 1.06 g/mL) is avaible. How many mL of KOH solution and water are needed? What is the m/m% of the formed solution if its density is 1.020 g/mL)? (720 mL KOH; 1780 mL water; 2.15
%)
78.) How much cc. hydrochloric acid is needed for the preparation of 200 mL 1.5 M HCl solution? The concentrated HCl solution is 35 m/m% and its density is 1.18 g/mL.
(26.5)
79.) What is the mol% composition of that ammonia solution 20 g of which diluted by 85 g water gives a 0.5 mol/kg molality solution?

(4.6)

80.) 1.5 L HCl solution of 8.00 mol/kg molality (d = 1.050 g/mL) and 2000 g of 18 m/m% solution were mixed. Calculate the m/m% and the mol% of the HCl in the formed solution. (18.8; 10.92)

81.) The concentration of a hydrochloric acid solution is 181 g/L, its density is 1.29 g/mL.
What is the Raoult’s concentration and the m/m% of that solution which made of 130 mL of the above HCl solution and 60 mL of a 22 m/m% (d = 1.11 g/mL) HCl solution?
(5.3; 16.22)
82.) From how much 12.5 m/m% solution should 56 kg of water be removed by distillation to have a 20 m/m% solution? (149.3)
83.) A sugar-refinery works up 300 ton of sugar-beet a day. From 100 kg of sugar-beet 130
L of diluted sugar solution is prepared with a density 1.103 g/mL, and 7.5 m/m% sugar content. This diluted solution is evaporated till the residue contains 90 m/m% solute.
How much water is nedded to evaporate a day?

(394.322 ton)

25

84.) How many g of water is required if 12 g CuSO4 · 5 H2O should be dissolved to form a saturated solution at 50°C? At 50 °C 100 g water dissolves 33.3 g of CuSO4.
(27.38)

26

5. CALCULATIONS CONNECTED TO TITRIMETRY

1.)

How many cm3 of 0.1 M HCl solution and 1.0 M HCl solution are required to neutralize
50.0 cm3 0.4 % NaOH solution?
First the molarity of NaOH is calculated:
100 cm3 NaOH solution contains

0.4 g NaOH

3

1000 cm NaOH solution contains

4.0 g NaOH

Since the molar mass of NaOH is 40 g/mol, the solution is 4/40 = 0.1 M

a) According to the reaction equation for the neutralisation of 1 mole of NaOH 1 mole of HCl is required. At equal molarity equal volumes contain equivalent amounts.

Thus, for neutralization of 50.0 cm3 0.1 NaOH solution, 50.0 cm3 0.1 HCl solution is required. b) The concentration of 1 M HCl is ten times higher than that of 0.1 M NaOH, and, therefore, 1/10

of

its

volume

contains

the

necessary

amount

of

HCl:

50.0/10 cm3 = 5.00 cm3.

Thus, for neutralization of 50.0 cm3 0.1 M NaOH solution, 5.00 cm3 1 M HCl solution is required. 2.)

For the neutralization of 168 mg NaHCO3, 19.68 cm3 0.1 M HCl solution is consumed.
Calculate the factor of the HCl solution.

theoretical consumption (cm 3 ) factor = actual consumption (cm 3 )
Actual consumption: 19.68 cm3
Theoretical consumption: How many cm3 would be consumed from an HCl solution of
0.1 M exactly?
The chemical reaction is the following:
HCl + NaHCO3 = NaCl + H2O + CO2
The molar mass of NaHCO3: MW = 84 g/mol
Calculate the molar amount of 168 mg of NaHCO3.

27

0.168
= 0.002 mole, thus 0.168 g (0.002 mol) NaHCO3 reacts with 0.002 mol of HCl.
84
If the concentration of the HCl solution is exactly 0.1M:
1000 cm3 0.1 M HCl contains

0.1 mol HCl
. 10 cm-3 mol HCl x cm 0.1 M HCl contains
2
1000 : x = 0,1 : 2.10-3
3

x = 20 cm3
From an exactly 0.1 M HCl solution 20 cm3 would be consumed for 168 mg of
NaHCO3, thus f =

20
= 1.016
19.68

The factor of the above HCl solution of 0.1 M is 1.016. This means that 1 cm3 of this

HCl solution equals 1.016 cm3 of exactly 0.1 M solution, or in other words its 1.016 times more concentrated than the nominal value.
3.)

Calculate the molarity and percentage concentrations of a KOH solution 10.0 cm3 of which consumes 9.82 cm3 0.05 M sulphuric acid (f = 1.012) test solution.
The molar mass of KOH: MW = 56 g/mol
Solution 1:

1000 cm3 0.05 M H2SO4 measures

5.6 g (= 0.1 mol) KOH

1 cm3 0.05 M H2SO4 measures

5.6 mg KOH

9.82 · 1.012 = 9.94 cm3 measures (9.94 · 5.6)

55.66 mg KOH

Thus, 10.00 cm3 titrated KOH solution contains 55.66 mg = 0.056 g KOH. From this, the m/v % = 0.56 % or the molarity =

0.56 ⋅ 10.00
= 0.1.
56

Solution 2:

Considering that in chemical reactions the amounts react with one another are determined by the stoichiometric equation (1 mol H2SO4 reacts with 2 moles of KOH), at the equivalence point the following relation is valid:
2 · vst · Mst · f = vu · Mu where: vst = the consumption of the standard solution in cm3
Mst = the molarity of the standard solution

28

f = the factor of the standard solution vu = the volume of the unknown solution
Mu = the molarity of the unknown solution
The molarity of the KOH solution: M u =

9.82 ⋅1.012 ⋅ 0.1
= 0 .1
10.0

Thus, the KOH solution is of 0.1 M and since the molar mass of KOH is 56g/mol:
0.1 ⋅ 56
= 0.56 % .
10

4.)

1 g solid KI is dissolved in sulphuric acid and 10 cm3 0.1 M H2O2 is added. Calculate the percentage of KI, which is reacted with hydrogen peroxide.
The amount of KI which is equivalent to 10 cm3 0.1 M H2O2 is to be calculated.
1000 cm3 0.1 M H2O2 contains

0.1 mol H2O2

10 cm3 contains

10-3 mol H2O2

10-3 mol H2O2 reacts with a double amount (2 · 10-3 mol) of KI:
2 KI + H2O2 + H2SO4 = K2SO4 + I2 + 2 H2O
Potassium iodide is a reducing agent since it is oxidized to iodine with the loss of one electron: 2 I- – 2e- = I2
1 mol KI

166 g

2 · 10-3 mol KI

xg

1 : 2 · 10-3 = 166 : x x = 166 · 2 · 10-3 = 0.332 g
Of 1 g KI 0.332 g reacted with 10 cm3 0.1 M H2O2 solution, thus, 33.2 % was consumed. 5.)

15 cm3 20 % hydrochloric acid solution (d = 1.10 g/cm3) is neutralized. How many cm3
0.5 mol/dm3 sodium hydroxide solution is consumed?

6.)

What is the concentration of that hydrochloric acid solution of which 10.00 cm3 can be neutralized by 11.50 cm3 NaOH solution of 0.1012 mol/dm3?

7.)

(180.8)

(0.1164)

What is the concentration of that NaOH solution of which 15.00 cm3 can be neutralized by 13.26 cm3 HCl solution of 0.15 mol/dm3 (f = 1.00)?

(0.1333)

29

8.)

1.2 dm3 standard state HCl gas is absorbed in 1.00 dm3 distilled water. What is the volume of that sodium hydroxide solution of 0.1 M (f = 0.998) which neutralizes 10.00 cm3 of the formed HCl solution?

9.)

(4.91)

15.00 cm3 HCl solution of 20 m/m% (density = 1.10 g/cm3) is neutralised. How many cm3 of 0.5 M NaOH solution is needed?

(180.8)

10.) Calculate the concentration of that sulfuric acid solution of which 10.00 cm3 can be neutralized by 12.45 cm3 KOH solution of 0.0963 mol/dm3. (0.0599)
11.) 20 cm3 sulphuric acid solution of an unknown concentration is neutralized with 12.14 cm3 0.10 mol/dm3 sodium hydroxide solution. Calculate the mol/dm3 concentrations of hydrogen ion and sulphuric acid as well.

(0.03; 0.06)

12.) Calculate the concentration of that phosphoric acid solution of which 20.00 cm3 can be neutralized by 23.05 cm3 NaOH solution of 0.0157 mol/dm3.

(0.0060)

13.) How many cm3 0.02 M KMnO4 solution are required to oxidize 0.500 g
[Fe(NH4)2(SO4)2 · 6 H2O] (Mohr's salt) in a strongly acidic medium, according to the incomplete equation: Fe2+ + MnO4- = Fe3+ + Mn2+? (12.75)

14.) Calculate the concentration of that oxalic acid solution of which 10.00 cm3 can be oxidized by 11.88 cm3 0.02 M (f = 0.999) KMnO4 solution in acidic medium. (0.0593)
15.) To 11.0 g of chlorine water KI is added and the liberated iodine consumes 11.0 cm3
0.2 M Na2S2O3 solution. Calculate the molality of the chlorine-water sample. (0.05)
16.) 20.00 cm3 of an I2 solution consumes 15.20 cm3 0.4 M Na2S2O3 standard solution.
Calculate the molarity and m/v % concentrations of the iodine solution.

(0.075; 1.90)

17.) 10.00 cm3 of an acetic acid solution with an unknown concentration is diluted to 100.00 cm3. From this stock-solution 10.00 cm3 samples are titrated with a 0.0849 mol/dm3
KOH solution. The average of the consumptions is 12.05 cm3.
What is the concentration of the stock-solution?

(0.1023)

Calculate the concentration of the original solution.

(1.0230)

How many g of acetic acid were in the original 10.00 cm3 sample?

(61.3827)

18.) For the titration of a formic acid solution 9.65 cm3 0.0935 mol/dm3 NaOH solution were consumed. How many cm3 of a 0.0200 mol/dm3 slightly basic KMnO4 solution are consumed by the same acid solution?

(37.07)

30

19.) To 10.00 cm3 of ammonia solution 20.00 cm3 0.1364mol/dm3 HCl solution is added, and the excess of the acid is titrated by a 0.0936 mol/dm3 NaOH solution. The average of the consumptions is 9.45 cm3. Calculate the molarity and the m/v% concentration of the ammonia solution.

(0.1843; 0.33)

20.) Calculate the hardness of that tap-water of which 50 cm3 11.43 cm3 0.0235 mol/dm3
EDTA solution were consumed.

(30.08 German degree)

31

6. ELECTROLYTIC DISSOCIATION

1.)

What are the concentrations of Ba2+ and OH- ions expressed in mol/dm3 unit in a solution, 500 cm3 of which contains 8.55 g Ba(OH)2, and α = 0.42?
Calculation of the molar concentration:
500 cm3 solution contains

8.55 g Ba(OH)2

1000 cm3 solution contains 17.1 g Ba(OH)2
Ba(OH)2 =

17,1
M

=

17,1
171

= 0.1 mol/dm3

Calculation of dissociated moles if α = 0.42: α= dissociate d moles total moles

Dissociated moles = α · total moles = 0.42 · 0.1 = 0.042
Thus, in 1 dm3 0.1 molar Ba(OH)2 solution 0.042 mol Ba(OH)2 dissociates.
The equation:

Ba2+ + 2 OH-

Ba(OH)2

Thus, the concentrations are:
[Ba2+] = 0.042 mol/dm3
[OH-] = 0.084 mol/dm3

2.)

Calculate the concentrations of H+ and CH3COO- ions and the degree of dissociation in an 0.10 M aqueous acetic acid solution if Ka = 1.86 · 10-5 at 5 °C?
The equilibrium equation:

H+ + CH3COO-

CH3COOH

According to the law of mass action:

[ H + ][CH 3COO − ]
K =
[CH 3COOH ]

Substituting the data; x = hydrogen ion concentration:

1, 86 ⋅ 10 −5 =

x2
0,1 − x

Rearranged to a quadratic equation: x2 + 1.86 · 10-5 x - 1.86 · 10-6 = 0.
32

x = 1.35 · 10-3.

The positive solution:

Thus, the ion concentrations are:
[H+] = [CH3COO-] = 1.35 · 10-3 mol/dm3
The degree of dissociation calculated from the [H+] = c0 · α relationship:

α = 1.35 · 10-2.
Notes:
B + C type equilibria, for 1 dm3 solution, the following is

a) In case of the A true: dissociated moles = α · initial moles = [B] = [C]
[A] = initial moles - dissociated moles = initial moles - α · initial moles =
= initial moles · (1- α)
If initial moles = c0 , then [A] = c0 · (1- α); and and [B] = [C] = α · c0

2
2
[ B ][C ] (α ⋅ c0 )(α ⋅ c0 ) = (αc0) = α c0
Ka =
=
c0(1 − α) 1 − α c0 (1 − α )
[ A]

B + C type equilibria, one may solve the above equation first

Thus, only for A

for α and then for the hydrogen ion concentration.
b) In most cases, the calculation may be simplified neglecting the x value in the denominator of the expression K =

x2
2
2
. Then c0 K ≅ x , and K ≅ α c0 . c0 − x

This simplification brings about less than 1 % error if α < 10-2, or

3.)

K
< 10-4. c0 The degree of dissociation of a 1.36 · 10-3 molar acetic acid solution is 0.11. Calculate the dissociation constant and pKa values.
The concentration of acetic acid: 1.36 · 10-3 M;
The equilibrium equation:

CH3COOH

The equilibrium concentrations:

α = 0.11
CH3COO- + H+

[H+] = 0.11 · 1.36 · 10-3 mol/dm3

[CH3COO-] = 0.11 · 1.36 · 10-3 mol/dm3
[CH3COOH] = 1.36 · 10-3 - 0.11 · 1.36 · 10-3 mol/dm3

33

The equilibrium constant:

Ka =

[1,36 ⋅10

−3

− 0,11 ⋅1,36 ⋅10

−3

= 1,85 ⋅10

−5

]

pKa = - lg · Ka = - lg · 1.85 · 10-5 = 5 - lg 1.85 = 4.73

Thus, the pKa value:
4.)

[0,11 ⋅1,36 ⋅10 −3 ] ⋅ [0,11 ⋅1,36 ⋅10 −3 ]

What are the ion concentrations (in mol/dm3) in a 6 · 10-3 molar MgCl2 solution? What is the ionic strength of the solution?
The equation of dissociation:

MgCl2 = Mg2+ + 2 Cl-

Thus, from 1 mol MgCl2, 1 mol Mg2+ and 2 mol Cl- form.
According to the complete dissociation, in a 6 · 10-3 molar solution the ion concentrations are:
[Mg2+] = 6 · 10-3 mol/dm3
[Cl-] = 12 · 10-3 mol/dm3
Thus, the ionic strength of the 6 · 10-3 molar MgCl2 solution is:
I=

5.)

Calculate the van't Hoff factors for the following electrolyte solutions if, in each case, α
= 1.

6.)

a) NaCl; b) Li2SO4; c) Al2(SO4)3?

(2; 3; 5)

How many undissociated molecules are there in 20 cm3 0.01 M acetic acid solution if α = 0.2?

7.)

6 ⋅ 10 − 3 ⋅ 4 + 12 ⋅ 10 − 3 ⋅ 1 0,024 + 0,012
=
= 0, 018
2
2

(9.64 · 1019)

By what factor does the degree of dissociation of 1 dm3 0.1 molar formic acid solution change if 0.05 mol sodium formate is added? (Ka = 1.77 · 10-4)

8.)

The degree of dissociation of an 0.375 molar hydrogen cyanide solution is
3.57 · 10-3 %. Calculate the dissociation constant and pKa values.

9.)

(11.4)

(4.8 · 10-10; 9.32)

Rewrite the list below in descending order of acid strength and calculate the Ka values.
Hypochlorous acid

pKa = 7.43

Periodic acid

pKa = 1.64

Nitrous acid

pKa = 3.40

Acetic acid

pKa = 4.73

34

10.) How many grams of arsenic are there in 1 dm3 arsenious acid (H3AsO3) solution in which the hydrogen ion concentration is found 9 · 10-6 mol/dm3? In aqueous solution, arsenious acid dissociates as a monobasic acid, Ka = 6 · 10-10.

(10.12)

11.) Calculate the molarity of an ammonia solution in which the degree of dissociation is
0.5%. (Kb = 1.8 · 10-5)

(0.72)

12.) Calculate the hydrogen ion concentration at 25 °C of an 0.2 molar formic acid solution
(Ka = 1.77 · 10-4).

(5.9.10-3)

13.) Calculate the hydrogen ion concentration at 25 °C of an 0.02 molar sulphurous acid solution. (Consider sulphurous acid as a monobasic one.) (Ka = 1.7 · 10-2) (1.18 · 10-2)
14.) What is the α value in a 2.0 · 10-2 molar NH3 solution? Calculate the dilution where the degree of dissociation of NH3 is 5%. (Kb = 1.79 · 10-5)

(2.95 · 10-2; 147)

15.) Calculate the molarity of a trichloroacetic acid solution in which the hydrogen ion concentration is 6.47 · 10-2 M and the concentration of the undissociated acid molecules is 3.235 · 10-2 M. Calculate the degree of dissociation and the dissociation constant as well. (9.7 · 10-2; 0.667; 1.29 · 10-10)

16.) In an 0.01 M HIO4 solution, the concentration of the anion is 7.6 · 10-3 mol/dm3.
Calculate the extent of dilution required to reach a 90% degree of dissociation.
(3.37)
17.) Calculate the ionic strength of the Ringer's solution. (The concentrations of NaCl, KCl,
CaCl2 and NaHCO3 are 0.85, 0.02, 0.02 and 0.01% respectively.)

(0.15)

18.) The pH of an unknown monobasic weak acid equal the pH of a 0.01 M HCl solution.
Calculate the degree of dissociation of the weak acid. What is the Ka of the weak acid?
(α = 0.0225; Ka = 0.00023)
19.) The pH of a 7.14 · 10-2 molar pyridine solution is 9.0. Calculate the base dissociation constant and the degree of dissociation of pyridine.

(0.00014; 1.4 · 10-9)

20.) 10.00 cm3 of an acetic acid solution of pH 3.0 can be neutralized by 5.63 cm3 of 0.1 molar (f = 0.987) NaOH solution. Calculate the degree of dissociation and the acid dissociation constant of acetic acid.

(0.018; 1.83 · 10-5)

21.) In a 0.010 molar periodic acid (HIO4) solution the concentration of the anion is 7.6·10-3.
At what factor of dilution is the degree of dissociation 90%? (Ka = 0.0241) (3.36)

35

22.) What are the pH and the degree of dissociation of acetic acid if 100 cm2 0.10 M acetic acid solution and 100 cm3 hydrochloric acid solution of pH 4 are mixed?
(3.01; 0.01846)

36

7. DILUTE SOLUTIONS

1.)

Calculate the osmotic pressure of a 10 m/m % aqueous saccharose (C12H22O11) solution at 20 °C. Calculate the freezing point and the boiling point of the solution as well. (d20 °C
= 1.050 g/cm3)

a)

Calculation of the osmotic pressure.
First the molarity of the solution is calculated:
The volume of 100 g 10 m/m % solution is:

100
= 95.2cm 3
1.050

95.2 cm3 solution contains 10 g saccharose
1000 cm3 solution contains x g saccharose
92.5 : 1000 = 10 : x x = 105
Thus, the solution is of 105 g/dm3 concentration. Correspondingly, its molarity is
105/MW = 105/342 = 0.31 mol/dm3.
According to the relation: π = cRT (R = 8.314 dm3 · kPa · K-1 · mol-1): π = cRT = 0.31 mol/dm3. 8.314 J K-1 mol-1 · 293 K = 753.86 kPa
Thus, the osmotic pressure of a 10 m/m % saccharose solution at 20 °C is 753.86 kPa.
b)

Calculation of the freezing point.
First the molality is calculated; according to the 10 m/m % percentage:
90 g solvent contains

10 g solute

1000 g solvent contains

x g solute

90 : 1000 = 10 : x x = 111 g
Thus, 1000 g solvent contains 111 g saccharose, i. e. the molality is 111/342 = 0.32 mol/kg solvent.
Calculation of the freezing-point depression.
Considering the molal freezing-point depression of water (Kf = 1.86 °C kg/mol):

37

Substituting the previously calculated molality value (m = 0.32 mol/kg):
∆Tf = cR · Kf = 0.32 · 1.86 = 0.59 °C
Since the fp. of the solvent is 0 °C, the solution freezes at -0.59 °C.
c.)

Calculation of the boiling-point elevation.
Substituting the previously calculated molality value (m = 0.32 mol/kg):
∆Tb = cR · Kb = 0.32 · 0.52 = 0.17 °C
Thus, the boiling-point elevation of the solution is 0.17 °C, i.e. the solution boils at
101.325 kPa pressure at 100.17 °C.

2.)

Calculate the osmotic pressure of a 6 · 10-3 molar MgCl2 solution at 20 °C. What would be the osmotic pressure if MgCl2 were a non-electrolyte?
In a 6 · 10-3 molar MgCl2 solution the sum of the concentrations of the dissociated particles is 6 · 10-3 + 12 · 10-3 = 1.8 · 10-2 mol/dm3. π = cRT = 1.8 · 10-2 mol/dm3 · 8.314 dm3 · kPa · K-1 · mol-1 · 293 K = 43.77 kPa
Thus, the osmotic pressure of a 6 · 10-3 molar MgCl2 solution at 20 °C is 43.77 kPa.
If MgCl2 is regarded as a non-electrolyte, the molarity = 6 · 10-3, i.e. one-third of the existing ion concentration. Thus, the osmotic pressure would be one-third of the previously calculated value:

43.77
= 14.59 k Pa
3
3.)

Calculate the osmotic pressure of a 2 % aqueous carbamide solution at room temperature. 4.)

(810.6 kPa)

Calculate the freezing-point depression and the boiling-point elevation of a 2.3 m/m % aqueous glucose solution. Kf = 1.86, Kb = 0.52.

5.)

Calculate the osmotic pressure of a 10 % aqueous hemoglobin solution at 20 °C.
(MW= 68,000 g/mol).

6.)

(3.643 kPa)

Calculate the percentage of a NaCl solution which is isotonic with a 5 % glucose (MW
= 180 g/mol) solution.

7.)

(0.24; 0.068)

(0.87)

Calculate the osmotic pressure of human blood at 20 °C considering that the percentage of the isotonic NaCl solution is 0.9.

(749.81)
38

8.)

Dissolving 0.241 g lactic acid (C3H6O3) in 53.5 g water (Kb = 0.52 °C · kg · mol-1), the boiling point is measured 100.0291 °C. Calculate the degree of dissociation according to the equation:
CH3CHOHCOO- + H+

CH3CHOHCOOH
9.)

(0.12)

Calculate the freezing point of a 60 g/dm3 K2SO4 solution of 1.050 g/cm3 density if the osmotic pressure of the solution at 0 °C is 1.62 MPa. Kf = 1.86

(-1.36)

10.) Calculate the osmotic pressure of a 4.0 m/m % CaCl2 solution (d = 1.032 g/cm3) at
100 °C if the boiling point of the solution is 100.51 °C at atmospheric pressure. (2.96)
11.) The atmospheric boiling point of a 6.5 m/m% AlCl3 solution is 101 °C and its density is
1.05 g/cm3. Calculate the chloride ion concentration of the solution and the osmotic pressure at 0 °C. Kb = 0.52.

(1.41; 4.31)

12.) Calculate the molar mass of the unknown compound, 2.320 g of which, dissolved in
18.27 cm3 water, exhibits 1.237 °C freezing-point depression. Kf = 1.86.

(191)

13.) Solutions of equal percentage are prepared from glucose (MW=180) and saccharose
(MW=342). Compare the osmotic pressures at a given temperature.
(The osmotic pressure of the glucose solution is higher by a factor of 342/180.)
14.) 1.76 g sulphur is dissolved in 27.5 g benzene. The freezing-point depression of the solution is measured 1.28 °C. Calculate the molecular state (the number of atoms in a sulphur molecule). Kf

(benzene)

= 5.12 °C · kg · mol-1, the atomic mass of sulphur = 32.

(8)
15.) 3.52 g of an unknown compound is dissolved in 23.8 g acetic acid. The boiling-point elevation of the solution is measured 1.68 °C. (Kb = 3.07 °C · kg · mol-1, α = 0.)
Calculate the molar mass of the unknown.

(270)

16.) Calculate the ebullioscopic constant of dioxane if the solution made of 173 g dioxane and 2.58 g naphthalene exhibits 0.364 °C boiling-point elevation.

(3.13)

17.) 8.15 g of an unknown compound is dissolved in 155 g water. The freezing point of the solution is measured -1.63 °C. Calculate the molar mass of the unknown. Kf = 1.86.
(60.33)

39

18.) 25 g glucose (C6H12O6) is dissolved in 215 g water. Calculate the freezing point of the solution. Kf = 1.86.

(-1.20)

19.) Calculate the freezing point of an aqueous urea solution if its osmotic pressure is 0.8
MPa at room temperature. Kf = 1.86. (-0.61)
20.) An ethanolic solution contains 4% aniline and 2.6 % nitrobenzene. Calculate the boiling point of the solution. Bp of ethanol = 78.3 °C, the molal bp-elevation value = 1.04 °C · kg · mol-1.

(79.01)

21.) Calculate the molarity of a glycerol solution with 1.02 g/cm3 density and freezing point
-2.00 °C. Kf = 1.86.

(1.0063)

22.) 1.755 g of an unknown organic compound is dissolved in 500 cm3 water and exhibits
0.1295 MPa osmotic pressure at 0 °C. Calculate the molar mass of the unknown.
(61.58)
23.) 0.0156 g protein is dissolved in water. The volume of the solution is 125 cm3. The osmotic pressure is measured 26 Pa at 25 °C. Calculate the molar mass of the protein.
(1.189 · 104)
24.) A patient suffers from a lack of 0.205 g potassium ion. This amount must be supplemented in an isotonic (0.3 Osmol/dm3) infusion with a [K+]/[Na+] ratio of 1:20.
How many grams of KCl and NaCl should be dissolved in water? What is the volume of the infusion? (0.392 g KCl; 6.143 g NaCl; 0.73 dm3)
25.) 25 g of an unknown sample is dissolved in 200 g of carbon tetrachloride. The boiling point of the solution is 77.4 °C. Calculate the molar mass of the unknown compound. Kf
= 5.02 °Ckg/mol, bpCCl4 = 76.8 °C.

(1045.8)

26.) Which of the following solutions has a higher osmotic pressure: the first one contains 5 g of naphthalene (C10H8), the second one contains 5 g of anthracene (C14H10) in 1 L of toluene. State your reason and calculate the osmotic pressures of the solutions at 25 °C.
(πC10H8 = 96.68 kPa > πC14H10 = 69.66 kPa)
27.) What is the ratio of the masses of formalin (CH2O) and glucose (C6H12O6) that generate the same osmotic pressure in equal volume of solutions at the same temperature?
(m1/m2 = 1/6)

40

28.) What is the molar weight of a protein 1 g of which dissolved in 100 mL of water the osmotic pressure is 253 Pa at 17 °C? The density of the solution is 1 g/cm3. (9.52 · 104)
29.) 1.52 g of thioacetamide (CH3CSNH2) is dissolved in 100 g of water. What is the freezing point of the solution? Kf = 1.86.

(-0.377)

30.) An aqueous solution of glucose (C6H12O6) freezes at -0.56 °C. What is the boiling point of this solution at atmospheric pressure? Kf = 1.86, Kb = 0.52.

(100.16)

31.) One part of glucose (C6H12O6) solution and three parts of urea (CO(NH2)2) solution were mixed. The osmotic pressure of the glucose solution was 283.7 kPa, that of the urea solution was 141.9 kPa at the same temperature. What is the osmotic pressure of the mixture?

(177.35)

32.) Ethylene glycol (C2H4(OH)2) is an anti-freeze additive in car radiators. How much ethylene glycol is to add into 946 g of water to prevent it from freezing untill -3 °C? Kf
= 1.86.

(94.6g)

33.) The molar weight of hemoglobin was determined by osmotic pressure measurement.
100 mL of solution contains 5.27 g of hemoglobin, its osmotic pressure is 7.83 kPa at
0°C. What is the molar weight of hemoglobin?

(1.53 · 103 g/mol)

34.) The osmotic pressure of a 0.1 m/m% protein solution is 2.133 kPa at 20 °C. What is the molar mass of the protein if the density of the solution is 1 g/cm3.

(1.14 · 103 g/mol)

35.) What is the osmotic pressure of the etheric solution which was made of 2.5 g of aniline
(C6H7N) and 75 g of ether? The density of the solution is 0.720 g/mL.

(567 kPa)

36.) Which one of the following aqueous solutions has a lower freezing point: 10 m/m% of methanol (CH3OH) solution and 10 m/m% ethanol (C2H5OH) solution?

(Methanol)

37.) How many g of sugar (C12H22O11) is to dissolve in 100 g of water
a.

to increase the boiling point of water by 0.1 °C?

(Kf = 1.86, Kb = 0.52)

b.

to decrease the freezing point of water by 1 °C?

(a. 6.58; b. 18.38)

38.) 5.70 g of an unknown organic compound was dissolved in 100 ml of chloroform. The freezing point depression is 2.38 °C. The density of chloroform is 1.50 g/cm3, its Kf value is 4.68 °C. According to the elemental analysis the compound contains 32.0 m/m% C, 6.60 m/m% H, 18.7 m/m% N and 42.7 m/m% O. Calculate the molar mass of the compound and determine its formula.

(75; C2H5NO2)

41

39.) A phenol (C6H5OH) containing benzene solution has an osmotic pressure of 0.2 MPa at
20 °C. What is the m/m percentage of the phenol if the density of the solution is 0.91 g/cm3? (0.85)
40.) A 0.1 molar aqueous NaHCO3 solution freezes at -0.38 °C. To how many particles does the sodium bicarbonate dissociate? d = 1.009 g/cm3. Kf = 1.86

(2)

41.) What is the ratio of the freezing point depressions of solutions containing equal percentage of glucose (C6H12O6) and sucrose (C12H22O11)?

(1.9)

42.) 1.5 g of urea (CO(NH2)2) was dissolved in 0.2 kg of solvent and a 2.75 °C freezing point depression was observed. What is the molal freezing point depression (Kf) of the solvent? (22)

43.) The aqueous solution of an unknown compound boils at 100.09 °C. If an additional
0.496 g of unknown compound is dissolved in 100 g of solution the new boiling point is
100.13 °C. What is the molar mass of the unknown compound?

(65.9)

44.) What is the boiling point of a solution made by mixing 200 cm3 5.5 m/m% glucose
(C6H12O6) solution (d = 1.15 g/cm3) and 525 g glucose solution of 29.3 kPa osmotic pressure at 20 °C?

(100.04)

45.) What is the freezing point of a solution wich contains 0.62 m/m% ethylene glycol
(C2H6O2) and 1.84 m/m% glycerol (C3H8O3)?

(-5.72 °C)

46.) Solutions of equal percentage are prepared from two compounds having molar masses
MW1 and MW2. What is the ratio of the boiling point elevations of the two solution?
(MW2/MW1)
47.) The aqueous solution of an unknown compound freezes at -0.310 °C. If 0.248 g of the compound is dissolved in 50 g of the above solution the new freezing point is -0.465 °C.
What is the molar mass of the compound?

(60.1)

Constants for the calculations related to aqueous solutions:
R = 8.314 J/molK,
Kfp = 1.86 °Ckg/mol, Kbp = 0.512 °Ckg/mol

42

8. HYDROGEN ION CONCENTRATION; pH, BUFFERS

1.)

Calculate the pH of an 0.05 molar HCl solution.
The definition of pH:

pH = -lg [H+]

The mol/dm3 concentration of hydrogen ion should be calculated. HCl is a strong acid, which dissociates completely in dilute aqueous solutions:
H+ + Cl-

HCl

Therefore, in an 0.05 molar solution, the hydrogen ion concentration is:
[H+] = 0.05 mol/dm3. pH = - lg 0.05 = - lg 5 · 10-2 = - (lg 5 + lg 10-2) = 2 - lg 5 = 1.30
Thus, in an 0.05 molar HCl solution, the pH = 1.30.
2.)

Calculate the pH of an 0.44 % NaOH solution at 25 °C.
First the mol/dm3 concentration is calculated:
100 cm3 solution contains

0.44 g NaOH

1000 cm3 solution contains

4.4 g NaOH

[ NaOH ] =

4.4
=
40

0.11 mol/dm3

NaOH is a strong base and in dilute solutions it dissociates completely:
NaOH

Na+ + OH-

Therefore, in an 0.11 mol/dm3 solution, the hydroxide ion concentration is:
[OH-]= 0.11 mol/dm3.
In the knowledge of [OH-], the pH may be calculated by either of the two methods:
Solution 1:
The ion product of water is the following:

[H+] · [OH-] = 10-14

In logarithmic form:

- lg [H+] + (- lg [OH-]) = - lg 10-14
i.e.

from which:

pH + pOH = 14

pH = 14 - pOH

43

Returning to our example: [OH-] = 0.11 mol/dm3 and pOH = - lg (1.1 · 10-1) = - lg 1.1 = 0.959

thus,

pH = 14 - 0.959 = 13.04

Solution 2.:
Knowing the [OH-], the [H+] may be calculated from the ion product of water:
+

[ H ]=

1 ⋅ 10 − 14
1.1 ⋅ 10

=

−1

10 ⋅ 10 − 15
1.1 ⋅ 10

−1

= 9.10 ⋅ 10 − 14 mol/dm 3

pH = -lg (9.10 · 10-14) = 14 - lg 9.10 = 13.04
Thus, the pH of an 0.44 % NaOH solution is 13.04.
3.)

Calculate the pH of an 0.148 % aqueous propionic acid solution.

(Ka = 1.22 · 10-5)

The propionic acid (CH3−CH2−COOH) is a weak acid, it dissociate partially in an equilibrium. Calculate

first the concentration of the solution. The molar mass of

propionic acid is 74 g/mol, thus its 0.148 m/v% solution correspond to 1.48 g/dm3.
74 g/dm3 solute means

1 M concentration

1.48 g/dm3solute means

x M concentration

x=

1.48 ⋅1
= 0.02 M
74

Applying the mass-action law for the dissociation of the weak acid:
CH3−CH2−COOH

CH3−CH2−COO- + H+

[ CH 3-CH 2 -COO-][H +]
Ka =
[CH 3 − CH 2 − COOH]
The aim is to calculate [H+]. For this reason, in addition to the Ka value, the equilibrium concentrations of propionic acid and propionate ion are to be calculated. From the chemical equation it is obvious that
[H+] = [CH3−CH2−COO-] and, according to the small value of Ka, [CH3−CH2−COOH] is practically equal to the initial concentration (0.02 M). i.e.

44

Ka =

[H +]2
[ CH 3-CH 2-COOH]

[H+]2 = Ka · [CH3−CH2−COOH]
[ H + ]= K a ⋅ [ CH 3 − CH 2 − COOH] = 1.22 ⋅ 10 − 5 ⋅ 2 ⋅ 10 − 2 = 2.44 ⋅ 10 − 7 p H= − lg[ H + ]= − lg 2.44 ⋅ 10 − 7 = 3.31

Thus, the pH of an 0.148 % propionic acid solution is 3.31.
4.)

Calculate the pH of 500 cm3 0.10 molar acetic acid solution in which 4.5 g sodium acetate is dissolved (Ka = 1.85 · 10-5).
The mass action law also valid for the acid-salt mixture:
CH3−CH2−COOH

CH3−CH2−COO- + H+

[ CH 3-CH 2 -COO-][H +]
Ka =
[CH 3 − CH 2 − COOH]
The hydrogen ion concentration of a buffer solution made of a weak acid and its salt can be expressed as shown below:
[H + ] = K a

Cacid
C salt

In logarithmic form:
− lg[ H + ] = − lg K a − lg

pH = pK a + lg

C acid
C salt

C salt
C acid

The latter expression is the Henderson-Hasselbalch equation, used to calculate the pH of buffer solutions.
For basic buffers:

pOH = pK b + lg

C salt
Cbase

In our case:

Ka = 1.85 · 10-5

and

pKa = 5 - lg 1.85 = 4.73

45

cacid = 0.1 mol/dm3

C salt =

4 .5 g
9
9
= 9 g / dm 3 = mol / dm 3 = mol / dm 3 = 0.11 mol / dm 3
3
500 cm
M
82

pH = 4,73 + lg

0.11
0 .1

pH = 4.73 + lg 1.1 = 4.73 + 0.041 pH = 4.77

Thus, the pH of the buffer solution is 4.77.
5.)

Calculate the buffer capacity of the above buffer solution against acid and base.
Calculation of the acid capacity:
According to the definition of the buffer capacity, the molar amount of the monovalent

strong acid (x) reducing the pH of 1 dm3 buffer solution by 1 pH unit are to be calculated. The pH of the acetate buffer:

pH = 4 , 73 + lg

0 , 11
0,1

(1)

The addition of x mol/dm3 acid leads to the following changes: csalt = 0.11 − x cacid = 0.1 + x
According to the changes:

pH − 1 = 4, 73 + lg

0,11 − x
0,1 + x

(2)

Equation (2) itself is suitable for the calculation, but the following method is simpler:
Subtract equation (2) from equation (1):

p H −(pH − 1)=4.73+lg

1 = lg

0.11
0.11 − x
−(4.73+lg
)
0.1
0.1+x

0.11
0.11 − x
− lg
0 .1
0.1+x

1 = lg

0.11(0.1 + x)
0.1(0.11 − x)

10 =

0.11(0.1 + x)
0.1(0.11 − x)

46

0.11 − x = 0.011 + 0.11x
1.11x = 0.099 x = 0.09

Thus, the pH decreases by one pH unit if 0.09 mol monovalent strong acid is added to
1000 cm3 buffer solution.
Calculation of base capacity:
Similarly to the former calculation, x mol/dm3 monovalent strong base increasing the

pH of 1 dm3 of the buffer by one pH unit is to be calculated.
After addition of x mol/dm3 base:

csalt = 0.11 + x cacid = 0.1 - x

The pH of the new system:

pH + 1 = 4 , 73 + lg

( 0 , 11 + x )
( 0,1 − x )

(3)

Subtracting equation (1) from equation (3): pH + 1 − pH = 4.73 + lg

1 = lg

(0.11 + x)
0.11
−(4.73 + lg
)
(0.1 − x)
0 .1

(0.11 + x)
0.11
− lg
(0.1 − x)
0 .1

1 = lg

(0.11 + x)⋅0.1
(0.1 − x)⋅0.11

10 =

(0.11 + x)⋅0.1
(0.1 − x)⋅0.11

x = 0.082 mol/dm3

Thus, the base capacity of the buffer is 0.082 mol/1000 cm3.
6.)

Calculate the pH of an 0.02 mol/dm3 HCl solution. How many H3O+ ions are there in
1 dm3 solution? (1.70; 1.204 · 1022)

7.)

The pH of a HNO3 solution is 3.0. Calculate the molarity and the g/dm3 concentration of the solution.

8.)

(10-3; 0.063)

The pH of a NaOH solution is 12.0. Calculate
a) the H3O+ ion concentration.

(10-12)

b) the OH- ion concentration.

(0.01)

47

c) the mol/dm3 concentration of the solution. (0.01)
9.)

Calculate the pH and pOH values at 25 °C in the solutions in which
a) the hydrogen ion concentration is 1.93 · 10-4 mol/dm3

(3.71; 10.29)

b) the hydroxide ion concentration is 3.48 · 10-9 mol/dm3

(5.53; 8.46)

c) the concentration of perchloric acid is 10.72 g/dm3

(0.97; 13.03)

d) the hydrogen bromide concentration is 0.0278 mol/dm3
e) the potassium hydroxide concentration is 1.808 g /dm

3

(1.56; 12.44)
(12.51; 1.49)

f) the calcium hydroxide concentration is 3.25 · 10-2 mol/dm3?

(12.81; 1.19)

10.) 500 cm3 solution is made from 0.2 g NaOH. Calculate the [OH-] and the pH.
(0.01 M; 12)
11.) Calculate the pH of an 0.005 molar H2SO4 solution. (2.0)
12.) The pH of a hydrochloric acid solution is 2.51. Calculate the concentration of the solution. (3.09 ·10-3)

13.) Calculate the pH of a NaOH solution at 25 °C and 0.42 m/m % concentration (d = 1.045 g/cm3). (13.04)

14.) Calculate the percentage and the molar concentrations of a sulphuric acid solution of pH
2.3.

(0.049; 0.005)

15.) How could you prepare 3.5 dm3 nitric acid solution of pH 2 from a 8.88 m/v% solution?
(24.8 cm3 solution should be diluted)
16.) Calculate the concentrations of OH- of a KOH solution of pH 8.80 in g/dm3 and mol/dm3 units.

(1.07 · 10-4;

6.31 · 10-6)

17.) How could you prepare 250 cm3 solution of pH 2.60 from 0.1 M hydrochloric acid solution? (6.25 cm3 1 M hydrochloric acid solution should be diluted to 250 cm3)
18.) The 18 m/m % solution of a strong monoacidic base is of 8.99 mol% and of
5.40 mol/dm3. Calculate the molar mass of the base and the density of the solution. How much water should be added to 1 cm3 of the solution to adjust the pH to 11.75?
(40; 1.20; 960.3)
19.) Calculate the pH of an 0.05 M acetic acid solution. (Ka = 1.86 · 10-5)

(3.02)

48

20.) Two HCl solutions of equal volume are mixed. The pH value of one is 1.00 that of the other one is 3.00. Calculate the pH value of the mixture.

(1.297)

21.) The pH of a 0.15 molar monoacidic weak acid (HA) is 2.35. Calculate the acid dissociation constant of the acid.
HA

H+ + A-

(1.37 . 10-4)

22.) The dissociation constant of cyanoacetic acid at 25 °C is 3.56 · 10-3. Calculate the pH of a 50 g/dm3 solution of it.
NC−CH2−COO- + H+

NC−CH2−COOH

(1.34)

23.) Calculate the degree of dissociation of acetic acid in the following solutions: 1 M, 0.1 M and 0.01 M acetic acid. Ka = 1.76 · 10-5.

(4.186 · 10-3; 1.327 · 10-2; 4.195 · 10-5)

24.) Calculate the pH of the 3 · 10-3 molar acetic acid solution. (Ka = 1.85 · 10-5 mol/dm3)
(3.63)
25.) Calculate the percentage of the dissociated molecules in 0.1 molar hydrogen fluoride solution of pH 2.24.

(5.75)

26.) The pH of the 0.02 molar monochloroacetic acid is 2.34. Calculate the dissociation constant. (1.35 · 10-3)

27.) How many grams of NaOH are there in 200 cm3 solution having an equal pH as an
0.1 mol/dm3 NH3 solution? (KNH3=1.79 · 10-5 mol/dm3)

(0.011)

28.) 150 cm3 0.5 molar acetic acid solution is diluted to 250 cm3. Calculate the pH values before and after dilution. (Ka = 1.85 · 10-5)

(2.52; 2.63)

29.) 10.0 g 90.0 m/m % and 20.0 g 10.0 m/m % acetic acid solutions are mixed and the volume is made up to 2 dm3. Calculate the pH of the final solution. (Ka = 1.85 · 10-5)
(2.62)
30.) 10.0 g one molal formic acid solution is diluted to 5.0 dm3. Calculate the pH of the diluted solution. (Ka = 1.77 · 10-4)

(3.3)

31.) In an iodic acid solution of pH 1.733 the concentration of the iodate ions is 9 times higher than that of the undissociated iodic acid. Calculate the dissociation constant and the degree of dissociation.

(0.1665; 0.9)

49

32.) In an 0.20 molar ammonia solution at 40 °C the molar concentration of ammonium ion is less than that of ammonia by a factor of 99. Calculate the pH and the dissociation constant. (The ion product of water at 40 °C is 3.8 · 10-14.)

(10.72; 2.02 · 10-5)

33.) By what factor should an 0.01 M acetic acid solution be diluted to decrease the pH by 1 unit? (Ka = 1.85 · 10-5)

(89.29)

34.) 800 cm3 HCl solution of pH 3.0 is mixed with 200 cm3 KOH solution of pH 12.
Calculate the pH of the mixture.

(11.08)

35.) The pH of a hydrochloric acid solution is 1.20. In 300 cm3 of this solution 0.115 g CaO is dissolved and the solution is made up to 500 cm3. What is the pH of the new solution?
(1.53)
36.) To 100 cm3 of a 10-2 molar HCl solution 10 cm3 of 0.1 molar NaOH solution is added.
What are the pH before and after the addition of NaOH?

(2.00; 7.00)

37.) Calculate the pH of the solution made by mixing 150 cm3 0.2 molar sulphuric acid solution with 350 cm3 0.1 molar sodium hydroxide solution. (1.30)
38.) To 20 cm3 2 molar HNO3 solution, 200 cm3 NaOH is added and the pH is measured 12.
Calculate the molarity of the NaOH solution. (0.211)
39.) An 0.01 mol/dm3 sulphuric acid solution is neutralized with KOH solution of pH 12.00.
Write chemical equation. Calculate the molarity of the salt formed in the reaction.
(0.0033)
40.) Calculate the pH of the solution made by mixing 10 cm3 5 % NaOH solution with 10 cm3 5 % H2SO4 solution, and making the volume up to 200 cm3.

(12.06)

41.) Calculate the pH of a solution made by mixing 1.0 dm3 5.0 m/m% NaOH (d = 1.054 g/cm3) solution with 1.0 dm3 4.0 m/m % HCl solution (d = 1.020 g/cm3), and the density of the mixture is 1.019 g/cm3.

(12.99)

42.) To each of 100 cm3 10-2 molar HCl solutions the following volumes of 10-1 molar
NaOH solution are added respectively: a) 0.0 cm3, b) 9.0 cm3, c) 9.9 cm3, d) 10.0 cm3,

e) 10.1 cm3, f) 11.0 cm3 and g) 20.0 cm3. Calculate the pH of the mixtures. Plot the pH values against cm3 added NaOH.
43.) What is the molar fraction of formic acid in a solution 10.0 g of which was diluted to
2.0 dm3 and the pH was measured to be 3.50? (Ka = 1.77 · 10-4)

(3.2 · 10-3)

50

44.) 10 cm3 88 m/m% formic acid solution was diluted to 2 dm3. What is the density of the
88 % solution if the pH of the diluted solution is 2.36? (Ka = 1.77 · 10-4)

(1.17)

45.) In 200 cm3 0.83 molar acetic acid solution (Ka = 1.85 · 10-5), 34 g crystalline sodium acetate, CH3COONa · 3 H2O, is dissolved and made up to 500 cm3. Calculate the pH of the buffer solution made in this way. (4.91)
46.) Calculate the pH of the following series of buffer solutions made by the addition of 40 cm3, 15 cm3, 10 cm3 and 2.5 cm3 0.1 NH3 solutions to 10 cm3 0.1 NH4Cl solutions respectively. (Kb = 1.79 · 10-5)

(9.85; 9.43; 9.25; 8.65)

47.) By what factor does the pH of 1 dm3 distilled water change if 0.4 g solid NaOH is dissolved in it? (5)
48.) Calculate the change in pH if 0.4 g solid NaOH is dissolved in a buffer solution made by mixing 0.5 dm3 2 molar acetic acid solution with 0.5 dm3 2 molar sodium acetate solution. (0.01)

49.) The pH of an aqueous solution containing equal moles of hydrogen cyanide and potassium cyanide is 9.32. Calculate the dissociation constant of hydrogen cyanide.
(4.8 · 10-10)
50.) How many grams of Na-propionate should be dissolved in 1 dm3 1 molar propionic acid solution to elevate the pH by 1.5 units? (Ka = 1.32 · 10-5)

(11)

51.) How many moles of sodium acetate are required to make 500 cm3 buffer solution of pH 8.89? (Ka = 1.8 · 10-5)

(0.0542)

52.) What is the ratio of CH3COOH / CH3COONa in that buffer solutions of which the pH are a.) 4.25; b.) 4.43; c.) 4.73; d.) 5.03; e.) 5.21, respectively? (3.03; 2; 1; 0.502; 0.332)
53.) Calculate the buffer capacities of pure water at 25 °C.

(10-6)

54.) 500 cm3 0.10 molar ammonium chloride and 1.0 dm3 0.01 molar ammonia solutions are mixed and made up to 2.0 dm3. Calculate the pH of the buffer.

(8.55)

55.) How many grams of acetic acid and sodium acetate are there in 1 dm3 buffer solution with pH 4.73 and 0.10 mol/dm3 acid capacity?

(7.32; 10.00)

56.) Calculate the molarity of an ammonium chloride solution 50 cm3 of which, when mixed with 25 cm3 0.8 % ammonia solution, shows pH 9.25. (pKb = 4.74)

(0.24)

51

57.) What are the pH and the pOH of a 0.0561 mol/dm3 nitric acid solution at 25 °C?
(1.25; 12.75)
58.) What is the pH of a 15 n/n% sulfurous acid solution at 25 °C?

(0.301)

59.) Calculate the molarity a nitric acid solution of a pH 3.65 at 25 °C.

(2.23 · 10-4)

60.) 0.7365 g NaOH is dissolved in 630 cm3 of water. What are the pH and the pOH of the solution at 25 °C?

(12.466; 1.534)

61.) What are the pH and the pOH of a 12 m/m% hydrochloric acid solution if the density is
1.060 g/cm3?

(1.458; 12.541)

62.) How many cm3 of a 0.0382 molar HCl neutralize 15.00 cm3 KOH solution of pH 12.34?
The titration is performed at 25 °C.

(8.6)

63.) What is the pOH of the solution made by the mixing of 145 g 14 m/m% KOH solution and 257 cm3 0.0154 mol/dm3 (d = 1.131 g/cm3) KOH solution? Both solutions are
25°C, the volume contraction is neglected.

(0.0235)

64.) By what factor should a sulfuric acid solution of pH 1.24 be diluted to increase the pH to 3.34?

(100)

65.) What are the pH and the pOH of a 0.2541 molar acetic acid solution at 25 °C?
(2.66; 11.34)
66.) Ammonia gas was introduced into 520 cm3 of water, the pH of the formed solution is
13.11. How many cm3 of standard state ammonia gas were absorbed by the water?
(1641.5)
67.) 1.000 g of a mixture composed by KOH and NaOH was dissolved in water and the solution was made up to 1.00 dm3. The pH of the diluted solution is 12.35 at 25°C.
Calculate the composition of the mixture in n/n% and m/m% units. (28.6; 36.0 KOH)
68.) To 100 cm3 HCl solution of pH 2.0 100 cm3 KOH solution was added. The pH of the formed solution is 3.0.
a. Calculate the concentration and the pH of the KOH solution.
b. What will be the pH if an additional 100 cm3 KOH solution is added to the 200 cm3 pH = 3 solution formed before?

(a. 0.008, 11.9; b. 11.3)

69.) What is the pH of the solution which contains acetic acid and sodium acetate, both of
0.01 molar concentration? (Ka = 1.8 ·10-5)

(4.74)

52

70.) 100 cm3 0.1000 molar phenol solution and 82 cm3 0.1000 molar NaOH solution are mixed. Calculate the pH of the forming solution. Ka = 1.3 · 10-10.

(10.49)

71.) How many moles of acetic acid and sodium acetate were there in the buffer solution of pH 4.40, 1 dm3 of which 20 mg NaOH was dissolved and the pH was changed by 0.60?
Ka = 1.85 · 10-5.

(0.00104 mol acid and 0.000469 mol salt)

72.) To 200 cm3 0.3 molar acetic acid solution 4.5 mg NaOH is added. What is the pH of the forming solution? Ka = 1.85 · 10-5.

(2.75)

53

9. HETEROGENEOUS EQUILIBRIA
(CRYSTALLIZATION, SOLUBILITY PRODUCT, PARTITION COEFFICIENT)

1.)

From 90 g KNO3, a saturated solution is made at 75 °C and then cooled to
20 °C. Calculate the m/m % of the cooled solution and the amount of deposited KNO3.
The solubility data of KNO3:
Temperature, °C

0

10

20

30

40

50

60

70

80

100

g/100 g solvent

13

21

32

46

64

85

110

138

169

246

250
200
150 g/100g 100
50
0
0

20

40

60

80

100

temperature

a) The solubility at 75 °C can be obtained from the solubility curve: 150 g/100 g water.
150 g KNO3 dissolves at 75 °C in
90 g KNO3 dissolves at 75 °C in

100 g water x g water

150 : 90 = 100 : x x= 9000
= 60 g water
150

Thus, 90 g KNO3 dissolves in 60 g water at 75 °C.

54

At 20 °C, the solubility of KNO3 is 32 g/100 g water.
100 g water dissolves at 20 °C

32 g KNO3

60 g water dissolves at 20 °C

x g KNO3

100 : 60 = 32 : x x= 1920
= 19.2 g KNO3
100

Thus, at 20 °C, 19.2 g KNO3 dissolves in 60 g water.
The amount of the deposited salt: 90 - 19.2 = 70.8 g
b) At 20 °C, 60 g water dissolves 19.2 g KNO3, i.e.: in (60 + 19.2 =) 79.2 g solution in 19.2 g solute

100.0 g solution

x g solute

79.2 : 100 = 19.2 : x x= 1920
= 24,2
79 , 2

Thus, the saturated solution of KNO3 at 20 °C is of 24.2 % by mass.
2.)

How many grams of CuSO4 · 5 H2O will crystallize if 200 g 40 m/m% CuSO4 solution is cooled to 30 °C? The saturated solution at 30 °C is of 20 m/m%.
M(CuSO4) = 159.6 g/mol
M(CuSO4 · 5 H2O) = 249.6 g/mol
The mass of the solution: 200 g
The mass of dissolved CuSO4:

m = 200 g · 40/100 = 80 g

The mass of deposited CuSO4 · 5 H2O = x x g CuSO4 · 5 H2O contains:

x ⋅159 , 6 g CuSO4
249 , 6

The mass of CuSO4 remained in solution: 80 -

x ⋅159 , 6
249 , 6

The mass of the solution: (200 - x) g
Since the concentration of the cooled solution is 20 %, the following relation exists:

55

(200 - x) . 0,2 = (80 -

x ⋅159 , 6
)
249 , 6

x = 91.11

Thus, 91.11 g CuSO4 · 5 H2O deposits.
3.)

Calculate the solubility product of silver phosphate if its solubility is 6.50 · 10-3 g/dm3 at room temperature.
The solubility of Ag3PO4:
6.50 ⋅ 10 − 3 g/dm 3 =
The dissociation process of Ag3PO4:

6.50 ⋅ 10 − 3
= 1.55 ⋅ 10 − 5 mol/dm 3
419

Ag3PO4

3 Ag+ + PO43-

The ion concentrations of the saturated solution:
[PO43-] = 1.55 · 10-5 mol/dm3
[Ag+] = 3 · (1.55 · 10-5) = 4.65 · 10-5 mol/dm3

Thus, the solubility product:
Ksp = [Ag+]3 [PO43-] = (4.65 · 10-5)3 · 1.55 · 10-5
Ksp = 1.57 · 10-18
4.)

The Ksp of AgBr is 6.4 · 10-13. How does the solubility of AgBr change in a 5 · 10-3 molar KBr solution?
First the solubility of AgBr in pure water is calculated. The equation
AgBr

Ag+ + Br-

suggests that in a saturated solution [Ag+] = [Br-] = x.
Thus:

Ksp = x · x x2 = 6.4 · 10-13 = 64 · 10-14 x = 8 · 10-7

Thus, in a saturated solution of AgBr at 20 °C, the Ag+ and Br- concentrations are 8 ·
10-7 mol /dm3.

56

Returning to the dissociation equation:
8 · 10-7 mol Br- and 8 · 10-7 mol Ag+ origin from the same amount of AgBr, therefore the solubility of AgBr is:
[AgBr] = 8 · 10-7 mol /dm3 = 188 · 8 · 10-7 g /dm3 = 1.5 · 10-4 g /dm3.
The solubility of AgBr in a 5 · 10-3 molar KBr solution is calculated:
Now, the Br- concentration is determined by the added electrolyte according to its complete dissociation:
K+ + Br-

KBr
Thus:

[Br-] = 5 · 10-3 mol/dm3
[Ag+] = x
Ksp = 6.4 · 10-13

x=

6.4 ⋅ 10 − 13
5 ⋅ 10

−3

= 1.28 ⋅ 10 − 10

Thus, in a 5 · 10-3 molar KBr solution, the Ag+ concentration is 1.28 · 10-10 mol/dm3.
Therefore:
[AgBr] = 188 · 1.28 · 10-10 g/dm3 = 2.4 · 10-8 g/dm3.

Thus, in 5 · 10-3 molar KBr solution, the solubility of AgBr is 2.4 · 10-8 g/dm3.
The solubility of AgBr in pure water is: 1.5 · 10-4 g/dm3.
That in 5 · 10-3 molar KBr solution is: 2.4 · 10-8 g/dm3.

1.5 ⋅ 10 − 4
2.4 ⋅ 10 − 8

=

15 ⋅ 10 − 5
2.4 ⋅ 10 − 8

= 6000

Thus, the solubility of AgBr in a 5 · 10-3 molar KBr solution is ca. 6000-fold smaller than in water.
5.)

Calculate the final concentration of a 2% aqueous iodine solution which has been extracted with an equal volume of carbon tetrachloride. (K = 86)
CCCl 4
CH 2O

= 86

57

Solution 1:
In case of equal volumes, after extraction, 1 part of I2 is retained in the aqueous phase and 86 parts are transferred into the organic phase (a total 87 parts of iodine):
87 parts

2%

1 part

x%

x = CH 2O =

2 ⋅1
≅ 0 , 02
87

Solution 2:
After extraction:

I2 concentration in CCl4:

x

I2 concentration in water:

(2-x)

x
= 86
2−x
x=

86 ⋅ 2
= 1, 98
87

(2 - x) = 0,02

Thus, after extraction, the I2 concentration of the aqueous phase is 0.02 %.
6.)

The task is the recrystallization of 40 g potassium nitrate. How many grams of water are required to form a saturated solution at 80 °C? How many grams of crystalline potassium nitrate can be recovered by cooling the solution to 10 °C (see the Table above on the solubility data of KNO3)?

7.)

(23.7; 35.03)

How many grams of potassium chloride can be crystallized from 500 g solution saturated at 100 °C and then cooled to 20 °C? (The solubility of KCl: at
100 °C: 56.05 g/100g water, at 20 °C: 34.37 g/100 g water.)

8.)

(69.48)

164 g potassium nitrate solution, saturated at 40 °C, is cooled to 20 °C. How many grams of water are to be added to avoid crystallization (see data above)?

9.)

(100)

132 g potassium nitrate solution, saturated at 20 °C, is heated to 40 °C. How many grams of KNO3 can be dissolved in the warm solution (see data above)?
(32)

58

10.) Calculate the ion concentrations in the saturated solutions of the precipitates listed below (see also Table 1).
a) Ca(OH)2; (18 °C)

(0.011; 0.022)

b) Ba(COO)2; (18 °C)

(3.4 · 10-3)

c) MgNH4PO4.(25 °C)

(6.3 · 10-5)

11.) Calculate the amount of solute (in grams and moles) dissolved in 750 cm3 saturated solution of Ca(OH)2 at 18 °C (see also Table 1).

(0.616; 8.32 · 10-3)

12.) 30 cm3 of water can dissolve 1.00 g silver chromate at 20 °C. What are the equilibrium concentrations and the solubility product?
(2.0 · 10-4 M Ag+; 10-4 M CrO42-; 4.00 · 10-12)
13.) Calculate the pH of the following saturated solutions:
a) 18 °C, Ca(OH)2; b) 20 °C, AgOH; c) 18 °C, Mn(OH)2; d) 25 °C, Cr(OH)3; e) 18 °C,
Fe(OH)3 (see also Tables 1 and 2).

(12.58; 10.26; 9.87; 6.58; 5.00)

14.) How many dm3 water are required to dissolve 1 g BaSO4 at 25 °C (see also Table 1)?
(429)
15.) How many cm3 water are required to dissolve 1.0 mmol silver iodide at 25 °C (see also
Table 1)?

(81650)

16.) How many g of bismuth sulfide dissolve in 1 m3 of water at 20 °C? (see also Table 1)
(8.02 · 10-5)
17.) How many g of silver bromide can be dissolved in 2.5 m3 of water at 25 °C?
(4.125 ·10-4)
18.) Calculate the concentration of a saturated PbCl2 solution in mol/dm3 at 18 °C (see also
Table 1).

(0.03)

19.) Will a precipitate of PbSO4 form at 25 °C if, to 10.0 cm3 10-4 molar Pb(NO3)2 solution
10.0 cm3 10-3 M H2SO4 solution is added (see also Table 1)?
(Will not because 1.25 · 10-8 < 1.44 · 10-8.)
20.) By what factor will the concentration of silver ions change if, into a saturated AgCl solution, dry HCl gas is introduced at 25 °C until the Cl- concentration is 0.03 mol/dm3
(see also Table 1)?

(2365)

59

21.) How many grams of K2CrO4 are to be added to 1 dm3 Ag2CrO4 solution saturated at 20
°C to diminish the silver ion concentration by the factor of 10 (see also Table 1)?
(1.95)
22.) 10.0 dm3 Pb3(PO4)2 solution, saturated at 20 °C, contains 1.38 mg dissolved salt. By what factor does the solubility decrease if 1.64 g Na3PO4 is added to the solution? (205)
23.) By what factor does the solubility of MgNH4PO4 change if it is dissolved in 0.01 molar
NH4Cl (see also Table 1)?

(12.6)

24.) Calculate the solubility products of the hydroxides listed below from the corresponding pH data measured at 20 °C:
a) saturated Mg(OH)2; pH = 10.346;

(5.5 · 10-12)

b) saturated Al(OH)3; pH = 10.512;

(3.72 · 10-15)

25.) What is the solubility product of aluminium hydroxide if the pH of its saturated solution is 10.51?

(3.7 · 10-15)

26.) Which of the following solution is more basic: Saturated Fe(OH)2 or saturated
Pb(OH)2? (see also Table 1)

(Pb(OH)2)

27.) How many milligrams of AgOH can be dissolved at 18 °C in 1 dm3 sodium hydroxide solution of pH 13.0 (see also Tables 1 and 2)?

(0.027)

28.) Calculate the pH of the solution, in 1 dm3 of which 0.10 g Ca(OH)2 can be dissolved at
18 °C (see also Table 2)?

(13.02)

29.) Calculate the pH of the saturated solution of Mg(OH)2 at 20 °C (see also Table 1).
(10.46)
30.) Calculate the pH of the saturated solution of Mg(OH)2 at 20 °C if in 1 dm3 solution of it
1.9 g MgCl2 is dissolved (see also Table 1). (9.39)
31.) Will a precipitate of Ag2SO4 form if, to 1 dm3 0.10 molar AgNO3 solution, 0.20 cm3
49 m/m% sulphuric acid solution of 1.380 g/cm3 density is added at 25 °C (see also
Table 1)?

(Will not, because 1.38 · 10-5 < 7.7 · 10-5)

32.) At what pH value does a precipitate form in a solution of each 10-3 molar Cr3+ and Mn2+ ions at 18 °C by gradual elevation of the pH? Calculate the pH value after stoichiometric hydroxide has been added (see also Table 1).

(5.15; 9.04)

60

33.) The pH of a saturated Mn(OH)2 solution is 9.63. By what amount does the mass of the solid being contact with 10.0 dm3 solution change if the pH is decreased to 9.00?
(-0.326)
34.) In 50 g of water some NH4NO3 is dissolved to form a solution saturated at 100 °C. How many g NH4NO3 precipitates if the solution is cooled down to 20 °C? At 20 °C 100 g water dissolves 192 g salt, at 100 °C 100 g water dissolves 871 g salt.

(339.5 g)

35.) How many g of Pb(NO3)2 dissolves in 20 g water of 60 °C? 100 g water dissolves 94.7 g of Pb(NO3)2 at 60 °C.

(19.0)

36.) The saturated solution of MgCl2 · 6 H2O is 35.3 m/m% at 20 °C and 37.9 m/m% at 60
°C. Calculate the solubility of MgCl2 · 6 H2O at both temperatures. (54.6; 61.0 g/100 water) 37.) The saturated solution of NaCl is 25.8 m/m% at 20 °C and 27.0 m/m% at 50 °C.
Calculate the solubility of NaCl at both temperatures.

(35.8; 36.5 g/100 g water)

38.) How many g of potassium chloride precipitates from 250 g of aqueous potassium chloride solution of 80 °C cooling it to 20 °C? (27.66)

61

Table 1. Solubility products of selected compounds
Compound

Temperature

Compound

Ksp

Temperature

Ksp

(°C)

(°C)
25

1.56 · 10-10

Fe(OH)3

18

3.80 · 10-38

50

1.32 · 10-9

Li2CO3

25

1.70 · 10-3

100

2.15 · 10-9

Mg(OH)2

25

5.50 · 10-12

AgBr

25

7.70 · 10-13

20

1.20 · 10-11

AgI

25

1.50 · 10-16

MgCO3

20

2.60 · 10-5

Ag3AsO3

25

6.50 · 10-12

MgNH4PO4

25

2.50 · 10-13

Ag2CO3

20

6.08 · 10-12

Mn(OH)2

18

4.00 · 10-14

25

6.15 · 10-12

PbCl2

18

1.00 · 10-4

Ag2CrO4

20

4.06 · 10-12

25

1.08 · 10-4

AgOH

20

1.52 · 10-8

PbI2

25

8.70 · 10-9

Ag3PO4

20

1.80 · 10-18

PbSO4

25

1.44 · 10-8

Ag2SO4

25

7.70 · 10-5

Ag4[Fe(CN)6]

20

1.50 · 10-41

Ag3[Fe(CN)6]

20

9.80 · 10

-26

Al(OH)3

25

3.70 · 10-15

Temperature

Kw

As2S5

18

1.05 · 10-30

°C

x 10-14

Ba(COO)2

18

1.62 · 10-7

0

0.133

BaSO4

25

1.00 · 10-10

18

0.580
0.676

25

2.13 · 10-16

20

Bi(OH)3

25

1.008

18

-6

40

3.800

50

5.960

100

59.00

AgCl

Ca(OH)2
Cr(OH)3

25
18

5.47 · 10
6.70 · 10

-31

5.40 · 10

-31

Table 2. The ion product of water at different temperatures

62

10. THERMOCHEMISTRY

1.)

Calculate the amount of heat required to transform 108 g ice of 0 °C to a vapour of 100
°C. The melting heat of ice is +6.03 kJ/mol, the heat of evaporation of water is
+40.74 kJ/mol and the specific heat of water (c) is +4.1868 J/g · °C.
108 g water =

108 mol water = 6 mol water
18

The conversion of ice to vapour involves three subsequent steps:
a) From ice of 0 °C to water of 0 °C:
∆H'change in state = 6 mol · 6.03 kJ/mol = 36.18 kJ
b) From ice of 0 °C to water of 100 °C:
∆H'' = c · m · ∆t = 4.1868 J/g · °C · 108 g · 100 °C = 45217 J = 45.22 kJ
c) From water of 100 °C to vapour of 100 °C:
∆H'''change in state = 6 mol · 40.74 kJ/mol = 244.44 kJ
∆H(total heat required) = ∆H'+∆H''+∆H'''
∆H = 36.18 kJ + 45.22 kJ + 244.44 kJ = 325.84 kJ

Thus, for the conversion of 108 g ice of 0 °C to a vapour of 100 °C, +325.84 kJ heat is required. 2.)

Calculate the heat of combustion of methanol in kJ/mol and kJ/kg units.
The reaction of combustion is the following:
CH3OH (l) + O2 (g) = CO2 (g) + 2 H2O (l)
The heat of formation values (∆Hf):
CH3OH (l) = - 201.3 kJ/mol
O2 (g)

=

0

CO2 (g)

=

-393.8 kJ/mol

H2O (l)

=

-286 kJ/mol

The solution is based on the following considerations: The starting materials are decomposed into their elements, and the heat required to this is equal, but of an opposite sign, to their heat of formation. The reaction products are formed from their elements, and the heat effect of these steps is equal to their heat of formation.

63

Thus, the heat of a reaction can be obtained by subtraction of the sum of the heat of formation of the starting materials from that of the products.
In this case:
- 393.8 + 2(- 286) - (- 201.3) = - 393.8 - 572.1 + 201.3 = - 965.8 + 201.3 = - 764.5 kJ

Thus, the heat of combustion of methanol is -764.55 kJ/mol, or:
− 764.5 kJ − 23.89 kJ
=
= − 23890 kJ / kg
32 g g 3.)

Calculate the heat evolved (or consumed) when 14.7 dm3 hydrogen chloride (at STP) is formed from its elements. Calculate the heat of formation of HCl.
The bond energies:

H−H: 436 kJ/mol.

Cl−Cl: 243 kJ/mol.

H−Cl: 432 kJ/mol.
H2 (g) + Cl2 (g) = 2 HCl (g)

∆H' = ?

According to the equation, one H-H bond and one Cl-Cl bond are broken and two H-Cl bonds are formed.
Thus, the heat of reaction:
∆H' = (436 + 243 - 2(432)) kJ = - 185 kJ

The heat of formation: ∆Hf(HCl) = ∆H'/2 = - 92.5 kJ/mol
Because

14,7 dm 3
= 0.6 mol HCl(g),
24.5 dm 3 / mol

Thus, ∆H = - 92.5 kJ/mol · 0.6 mol = - 55.5 kJ

Thus, formation of 0.6 mol HCl (g) is accompanied with the liberation of 55.5 kJ heat.
4.)

In the reactive engine of the rocket Saturn V, kerosene (ca. C12H26) is combusted with liquid oxygen. Calculate the energy produced in the engine supposing that 550 metric tons kerosene is combusted. (The heat of combustion of kerosene: ∆H = - 7510 kJ/mol.)
(2.43 · 1010 kJ = 2.43 · 104 GJ)

5.)

Chloromethane is synthesized by chlorination of methane. Write equation and calculate the reaction heat. The bond energies are:
C−H: 414 kJ/mol

C−Cl: 328 kJ/mol

Cl−Cl: 243 kJ/mol

H−Cl: 432 kJ/mol

(- 103)

64

6.)

Complete and balance the following word equations and calculate the reaction heats.

a) ethene + hydrogen;

b) ethyne + hydrogen;

c) propene + hydrogen;

d) butadiene + 1 mol hydrogen.

The bond energies are:
C−C: 344 kJ/mol.

C=C: 615 kJ/mol.

C−H: 414 kJ/mol.

C≡C: 812 kJ/mol

H−H: 436 kJ/mol.

(a,c,d: - 121; b: - 195)
7.)

Calculate the bond energy of elementary nitrogen. The following data are known:
Bond energies:
H−H: 436 kJ/mol and N−H: 391 kJ/mol
N2 + 3 H2

8.)

2 NH3

∆H = - 92 kJ.

(946)

Into 100 g water of 80 °C, 20 g ice of 0 °C is placed. Calculate the temperature after the thermal equilibration. (The heat of melting of ice: + 6.03 kJ/mol.)

9.)

(53.3)

15 g aluminium is combusted in a calorimeter to produce - 442.54 kJ heat. Calculate the heat of formation of aluminium oxide.

(-1591)

10.) Calculate the heat of formation of benzene if the combustion of 10 g benzene liberates
418.68 kJ energy. (The heat of formation of CO2 = - 393.98 kJ/mol; that of water = -286 kJ/mol.) (44.38)

11.) The heat of formation of glucose is -1257.71 kJ/mol, while that of water is -285.96 kJ/mol and that of carbon dioxide is -393.98 kJ/mol. Calculate the heat evolved during aerobe oxidation of glucose.
12.) Calculate the heat of the 3 C2H2

(-2822)
C6H6. The heat of combustion of acetylene is -

50.03 kJ/g, and that of benzene is - 41.87 kJ/g.

(-636.8)

13.) Calculate the change in temperature if 1 dm3 water dissolves 0.746 g KCl. (The lattice energy of KCl = 703.38 kJ/mol. The hydration energy of potassium ion is - 330.76 kJ/mol and that of chloride ion = - 347.50 kJ/mol.)

(0.06)

65

11. ELECTROCHEMISTRY

1.)

Calculate the electromotive force (e.m.f.) of the following galvanic cell at
25 °C: Zn | 0.15 mol/dm3 ZnSO4 solution || 0.32 mol/dm3 CuSO4 solution | Cu.
Determine the poles (ε°Cu = + 0.34 V; ε°Zn = - 0.76 V).
The potential of the copper electrode:

ε Cu = ε o +
Cu

0.059
⋅ lg[ Cu 2 + ]
2

ε Cu = 0.34 +

0.059
⋅ lg 0.32
2

εCu = 0.34 + (- 0.0146) = 0.325
That of the zinc electrode:

ε Zn = ε o +
Zn

0. 059
⋅ lg[ Zn 2 + ]
2

ε Zn = − 0.76 +

0.059
⋅ lg 0.15
2

εZn = - 0.76 + (- 0.0243) = - 0.784
E = εCu - εZn = 0.325 - (- 0.784) = 1.11 V

The e.m.f. of the cell:

Thus, the e.m.f. is 1.11 V. The negative pole is the zinc electrode and the positive one is the copper electrode.
2.)

CuCl2 solution is electrolyzed between platinum electrodes. How many grams of copper and how many dm3 of chlorine gas at STP form on the electrodes if 19,300 coulombs are passed through the solution?
96,500 coulomb produce
19,300 coulomb

63.5
2

g Cu

x

g Cu

96500 : 19300 = 63.5/2 : x

x=

63.5 ⋅19300
= 6.35 g copper
2 ⋅ 96500

66

6500 coulomb produce

22.41 dm3 Cl2 gas
2

19300 coulomb produce

x

dm3 Cl2 gas

96500 : 19300 = 22,41/2 : x

x=

22.41 ⋅19300
= 2.24 dm 3 Cl2
2 ⋅ 96500

Thus, when 19,300 coulombs passed through, the cell produces 2.24 dm3 Cl2 gas and
6.35 g Cu.
3.)

Will a redox reaction afford metal deposition, when immersing

a) a silver plate into FeSO4 solution,
b) an iron plate into AgNO3 solution,
c) an iron plate into SnCl4 solution,
d) an iron plate into SbCl3 solution,
e) a copper plate into HgCl2 solution,
f) a tin plate into CdSO4 solution,
g) a tin plate into SbCl3 solution?
4.)

Calculate the potential of a hydrogen electrode of 0.1 MPa at 25 °C immersed into the solutions listed below:

a) 0.01 molar HCl solution,
b) 0.01 molar NaOH solution,
c) 0.01 molar HF solution (Ka = 7.2 · 10-4),
d) 0.01 molar NH3 solution (Kb = 1.79 · 10-5),
e) 0.01 molar NaCl solution,
f) acetate buffer solution of 1:1 molar ratio (Ka = 1.86 · 10-5),
g) saturated Mg(OH)2 solution (Ksp = 5.5 · 10-12).
5.)

Calculate the potential of a copper electrode at 25 °C dipped into 300 cm3 solution containing 2 g CuCl2? (ε° = + 0.34 V)

6.)

(0.302)

Calculate the e.m.f. of a galvanic cell of Zn | 1 M ZnSO4 || 0.5 M CuSO4 | Cu composition. The standard potentials are: Zn2+/Zn : - 0.76 V, Cu2+/Cu : + 0.34 V.
(1.91)

67

7.)

Give an explanation of the function and calculate the e.m.f. for the following galvanic cell: Ni | 0.1 M NiSO4 || 0.1 M HCl | Pt (H2). (ε°Ni = - 0.23 V)

8.)

(0.201)

Explain the function and calculate the e.m.f. of the following galvanic cell:
Pt | 0.1 M FeCl2, 0.01 M FeCl3 || 1.18 M HCl | Pt (H2) at 25 °C. (In the Fe3+/Fe2+ redox system ε° = + 0.76 V.)

9.)

(0.701)

Calculate the Sn4+/Sn2+ concentration ratio in the galvanic cell of the following composition: Pt | SnCl2, SnCl4 || 0.2 M FeSO4, 0.3 M Fe2(SO4)3 | Pt, if the measured
e.m.f. at 25 °C is 0.532 V. (In the Sn4+/Sn2+ system ε° = + 0.1 V)

(4.92 · 104)

10.) Calculate the nickel ion concentration in a galvanic cell indicated below, in which the
e.m.f. is measured 1.043 V. Ni | NiSO4 || 7.17 · 10-2 mol/dm3 Ag+ | Ag. (1.74 · 10-3 )
11.) Calculate the dissociation constant of trichloroacetic acid if the e.m.f. of the cell below is measured 72 mV.
(Pt)H2 | 3 · 10-2 molar Cl3CCOOH || HCl solution of pH 1.0 | H2(Pt) (1.52 · 10-3)
12.) Calculate the pH in the trichloroacetic acid solution of the concentration cell noted below if the measured e.m.f. = 100 mV.
Pt(H2) | 0.1 M HCl || Cl3CCOOH | Pt(H2)

(2.69)

13.) Calculate the pH of the anodic solution of the concentration cell below if the measured
e.m.f. = 84 mV.
(Pt)H2 | unknown solution || solution of pH 3.41 | H2(Pt)

(4.83)

14.) Calculate the solubility product of silver chloride if the e.m.f. of the following cell is measured 0.23 V:
Ag | saturated AgCl + 1 M KCl || 0.1 M Ag+ solution | Ag

(1.60 · 10-10)

15.) A solution of a salt of a trivalent metal is electrolyzed for 40 min at a current of 2 A, and 1.4744 g metal is deposited. Calculate the atomic mass of the metal.

(88.92)

16.) 800 cm3 of exploding gas of 100 kPa and 18 °C is produced during the electrolysis of an acid solution for 25 min. Calculate the current.

(2.8)

17.) The total amount of chlorine gas is liberated from 1 dm3 of an 0.05 molar CaCl2 solution at a current of 2 A. Calculate the time required. (The efficiency of the electrolysis is 100 %.)

(80 min)

68

18.) An aqueous solution of NaCl is electrolyzed for 24 hours between mercury and graphite electrodes at 150 kA current. Write down the cathode and the anode processes.
Supposing a 100 % current efficiency, calculate the volume of the gas evolved at the anode at STP.

(1645 dm3)

19.) A dilute sodium hydroxide solution is electrolyzed for 15.0 min at 1.70 A current. How many dm3 of exploding gas are evolved at 25 °C and 93.4 kPa?

(0.315)

20.) Calculate the pH value of the solution of a hydrogen electrode connected to a standard hydrogen electrode when 177 mV e.m.f. is measured. (2.95)
21.) What can be the standard potential of the other electrode in that standard galvanic cell which contains a standard Mn2+/Mn electrode and its electromotive force is 0.63 V? (0.40 V Cd2+/Cd or – 1.66 V Al3+/Al)
22.) To 250 cm3 lead(II)nitrate solution of 0.100 M 1.60 g of metal is placed. How does the mass of the solid phase change if the metal is zinc, copper or nickel? (Zn: + 3.48g; Cu:
0; Ni: + 3.71g)
23.) To 100 cm3 copper(II)sulfate solution of 5.00 M an iron stick is immersed. What is the concentration of copper(II) and iron(II) ions after removing the metal stick if its mass increased by 770 mg? (the change of volume of the solution is neglected)
(1,0 M Fe2+; 4.0 M Cu2+)
24.) Into 100 cm3 10 m/m% copper(II) sulfate solution of 1.117 g/cm3 density 10.0 g iron powder is strewed. What will be the mass of the solid phase after the reaction? Calculate the m/m% composition of the solution after the reaction as well.
(10.54 g; 9.56 m/m%)
25.) How many g of copper scobs are to be placed into 100 g 10.0 m/m% silver nitrate solution to decrease the concentration of silver to the half?

(1.87)

26.) How many g of iron scobs are to be placed into 100 g 20.0 m/m% copper(II) sulfate solution to decrease the concentration of copper to 6.0 m/m%?

(4.91)

27.) An aqueous solution of silver nitrate is electrolyzed for 2 hours between graphite electrodes at 5 A current. Supposing a 100 % current efficiency, calculate the mass of precipitated silver and the volume of the gas evolved at the other electrode at STP.
(40.25 g; 2.08 dm3)

69

28.) 10.0 copper is to be prepared by the electrolysis of a copper(II) sulfate solution. How much time is required for the electrolysis at 5 A current?

(6079 s)

29.) What is the oxidation number of titanium in that compound from which 4.4760 g titanium is produced during 1 hour of electrolysis at 10 A current?

(4)

30.) The following galvanic cell was prepared at 25 °C:
+ X(sz) |Xn+(aq) || 1 M H+ (aq) | H2 (Pt) 101 kPa The electromotive force of the galvanic cell is 0.799 V. What is the electode potential of
X electrode?

(+ 0.799)

70