...Acid-Base Titration: Determination of the Percentages (%) of Sodium Carbonate (Na2Co3) and Sodium Hydroxide (Naoh) in a Mixture Title Acid-base titration: Determination of the percentages (%) of sodium carbonate (Na2CO3) and sodium hydroxide (NaOH) in a mixture Objective To determine the respective weight per cent of sodium carbonate and sodium hydroxide in a mixture by acid-base titration. Result and calculation Part A Titration 1 Titration number 1 2 3 Initial volume of burette( cm3) 5.10 2.70 9.70 Final volume of burette (cm3) 34.40 31.80 39.20 Total volume of HCl used (cm3) 29.30 29.10 29.50 Average volume of HCl required for titration =(29.30+29.10+29.50)/3 cm3 = 29.30 cm3 Titration 2 Titration number 1 2 3 Initial volume of burette( cm3) 4.50 14.00 2.70 Final volume of burette (cm3) 25.00 21.70 22.80 Total volume of HCl used (cm3) 20.50 20.30 20.10 Average volume of HCl required for titration =(20.50+20.30+20.10)/3 cm3 = 20.30 cm3 Part B Titration number Rough 1 2 3 Initial volume of burette( cm3) 4.9 4.80 3.60 2.20 Final volume of burette (cm3) 28.3 28.90 27.70 26.20 Total volume of HCl used until phenolphthalein decolourised (cm3) , x 23.4 24.10 24.10 24.00 Initial volume of burette after adding methyl orange indicator ( cm3) 28.3 28.90 27.70 26.20 Final volume of burette (cm3) 34.1 33.40 32.10 30.70 Total volume of HCl used until phenolphthalein decolourised (cm3) , y 5...
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...(4) Acid-base Titration using Method of Double Indicators Student Handout Purposes To determine the composition of the following mixture by double indicator method: 1. NaOH(aq) and Na2CO3(aq) 2. NaHCO3(aq) and Na2CO3(aq) Introduction Consider a mixture of NaOH(aq) and Na2CO3(aq). Reaction between HCl(aq) and Na2CO3(aq) takes place in two stages: HCl(aq) + Na2CO3(aq) ⎯→ NaHCO3(aq) + H2O(l) …………………. (1) HCl(aq) + NaHCO3(aq) ⎯→ NaCl(aq) + CO2(g) + H2O(l) …………. (2) While that between HCl(aq) and NaOH(aq) completes in only one step: HCl(aq) + NaOH(aq) ⎯→ NaCl(aq) + H2O(l) ……………….………. (3) Solution mixture of reaction (1) at the equivalence point is alkaline, that of reaction (2) is acidic and that of reaction (3) is neutral. Thus the whole titration should have three breaks in the pH curve, corresponding to the above three stages. Reactions (1) and (3) can be indicated by phenolphthalein and that of reaction (2) can be indicated by methyl orange. Stoichiometry confines each of the above pH reactions to react according to a mole ratio of 1 : 1. This means, say from equation (2), the number of mole of HCl(aq) determined from the methyl orange titration is equal to the number of mole of NaHCO3(aq). Likewise, total number of moles of NaOH(aq) and Na2CO3(aq) in the solution mixture can be calculated according to the volumes of HCl(aq) added at the end point Vol. of HCl indicated by the colour change of the phenolphthalein indicator. Alternatively, the Fig. 1: Titration curve for a mixture...
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...EXPERIMENT 5 REDOX TITRATION: TITRATION USING SODIUM THIOSULPHATE Objectives 1. To prepare a standard solution of potassium iodate for use to determine the concentration of sodium thiosulphate solution accurately. 2. To acquire the proper techniques of carrying out a titration. Introduction Redox titrations using sodium thiosulphate as a reducing agent is known as iodometric titration since it is used specifically to titrate iodine. The reaction involved is: I2 + 2Na2S2O3 2NaI + Na2S4O6 I2 + 2S2O32- 2I- + S4O62- In this equation I2 has been reduced to I- :2S2O32- S4O62- + 2e I2 + 2e 2I- The iodine/thiosulphate titration is a general method for determining the concentration of an oxidising agent solution. A known volume of an oxidising agent is added into an excess solution of acidified potassium iodide. The reaction will release iodine:Example: (a) With KMnO4 2MnO4- + 16H+ + 10I- 2Mn2+ + 5I2 + 8H2O (b) With KIO3 IO3- + 5I- + 6H+ 3I2 + 3H2O The iodine that is released is titrated against a standard thiosulphate solution. From the stoichiometry of the reaction, the amount of iodine can be determined and from this, the concentration of the oxidising agent which released the iodine, can be calculated. In an iodometric titration, a starch solution is used as an indicator as it can absorb the iodine that is released. This absorption will cause the solution to change to a dark blue colour. When this dark blue...
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...Titrations Practice Worksheet Find the requested quantities in the following problems: 1) If it takes 54 mL of 0.1 M NaOH to neutralize 125 mL of an HCl solution, what is the concentration of the HCl? 2) If it takes 25 mL of 0.05 M HCl to neutralize 345 mL of NaOH solution, what is the concentration of the NaOH solution? 3) If it takes 50 mL of 0.5 M KOH solution to completely neutralize 125 mL of sulfuric acid solution (H2SO4), what is the concentration of the H2SO4 solution? 4) Can I titrate a solution of unknown concentration with another solution of unknown concentration and still get a meaningful answer? Explain your answer in a few sentences. 5) Explain the difference between an endpoint and equivalence point in a titration. Solutions to the Titrations Practice Worksheet For questions 1 and 2, the units for your final answer should be “M”, or “molar”, because you’re trying to find the molarity of the acid or base solution. To solve these problems, use M1V1 = M2V2 . 1) 0.043 M HCl 2) 0.0036 M NaOH For problem 3, you need to divide your final answer by two, because H2SO4 is a diprotic acid, meaning that there are two acidic hydrogens that need to be neutralized during the titration. As a result, it takes twice as much base to neutralize it, making the concentration of the acid appear twice as large as it really is. 3) 0.1 M H2SO4 4) You cannot do a titration without knowing the molarity of at least...
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...CHEMISTRY: MATTER AND EQUILIBRIUM Indigestion and Titration: An Acid-Base Titration Imagine yourself as the Lead Analytical Chemist at Kaplan Industries. Your first big assignment is to investigate the strength of several commercial antacids for the Food and Drug Administration (FDA). They have sent five antacids to be tested with a back-titration that works as follows: • • • First, each antacid tablet is mixed with 40 mL of 0.1 M HCl—this acidic solution is the same stuff that is in stomach acid, and one antacid pill is nowhere near enough to neutralize all 40 mL of the acid. So, to see how much extra help each antacid pill needs to neutralize 40 mL of 0.1 M HCL, you add 0.05 M NaOH drop-by-drop to back-titrate the solution until the pH is neutral. What this means is that, the stronger the antacid tablet, the less NaOH it will take to help bring the acid to neutral. (In other words, the stronger antacid tablets counteract more of the original HCl, leaving the solution closer to neutral before the NaOH is added.) Here are your results: Maalox Mass of one dose antacid mL NaOH used in backtitration 20.0 g Tums 21.0 g Mylanta 18.0 g CVS brand 18.3 g Rennies 17.5 g 24.1 mL 22.4 mL 20.0 mL 19.9 mL 24.4 mL 1. Which is the strongest antacid, on a single-dose basis? Which is the weakest? Explain and show your calculations. 2. Which are the strongest and weakest, on a by-weight (mass) basis? 3. When people do back titrations, they usually watch the solution for a color...
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...Elijah Kim Mrs. Dobler Course 2 2/10/15 Titration lab Titration is used to find the specific amount of a standard solution in an amount of unknown liquid. Titration in this lab is performed with the chemical reactions between acids and liquids. During titration, we have to stop at the point where stoichiometric amounts of acids and bases are reacting. We can find that point by looking for indicators, for example in this lab the indicators were changes in color. So the purpose of all this is to determine the concentration of some acid solutions. 1. Measure 10 mL of your acid solution using graduated cylinder and and it to the Erlenmeyer flask 2. Add about 25 mL of distilled water to the flask 3. Place your Erlenmeyer flask under the buret and a white piece of paper under you Erlenmeyer flask 4. Record the initial buret reading 5. Start adding the NaOH to the flask dropwise, stopping at a faint pink point. 6. Record final buret reading 7. Repeat for a second and third trial Trial one Trial two Trial three Volume of acid sample: 10 mL 10 mL 10 mL Final Buret reading: 20.22 29.8 40.1 Initial Buret reading: 10.63 20.22 29.8 Net volume NaOH used: 9.6 9.6 10.3 Calculations Trial one Trial two Trial three Moles NaOH reacted: ...
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...Elijah Kim Mrs. Dobler Course 2 2/10/15 Titration lab Titration is used to find the specific amount of a standard solution in an amount of unknown liquid. Titration in this lab is performed with the chemical reactions between acids and liquids. During titration, we have to stop at the point where stoichiometric amounts of acids and bases are reacting. We can find that point by looking for indicators, for example in this lab the indicators were changes in color. So the purpose of all this is to determine the concentration of some acid solutions. 1. Measure 10 mL of your acid solution using graduated cylinder and and it to the Erlenmeyer flask 2. Add about 25 mL of distilled water to the flask 3. Place your Erlenmeyer flask under the buret and a white piece of paper under you Erlenmeyer flask 4. Record the initial buret reading 5. Start adding the NaOH to the flask dropwise, stopping at a faint pink point. 6. Record final buret reading 7. Repeat for a second and third trial Trial one Trial two Trial three Volume of acid sample: 10 mL 10 mL 10 mL Final Buret reading: 20.22 29.8 40.1 Initial Buret reading: 10.63 20.22 29.8 Net volume NaOH used: 9.6 9.6 10.3 Calculations Trial one Trial two Trial three Moles NaOH reacted: ...
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...Introduction Tissue culture, is the removal of cells from an animal or a plant, and growing the cells in a favourable artificial condition (Wiley, J.M et al, 2013). There are three types of cell culture: primary, this is cultivating the cells away from the parental tissue, secondary, which is when a primary culture is sub-cultured and cell lines, which can be continuous or finite depending on the life span of the culture. Virus titrations are used to estimate the virus concentration; it is a viral quantification technique. When detecting the virus, the cytopathic effect is looked at, whether there is lysis of the cells, vacuolation, formation of syncytia and the presence of inclusion bodies. TCID50, is the measure of the infectious titre. The end point dilution assay quantifies the amount of virus that is required to kill 50% of infected hosts or to produce a cytopathic effect in 50% of inoculated tissue culture cells (Kumar P, 2013). The purpose of the virus titration within tissue culture is to isolate and identify viruses within clinical samples, to carry out research on the viral structure, replication, genetics of the virus and the effect on the host cells, and also to prepare viruses for vaccine production. Results Table 1 Data used to determine the 50% endpoint using the Reed-Muench method Log of virus dilution | Infected test units | Cumulative infected (A) | Cumulative non-infected (B) | Ratio A/(A+B) | Percentage infected (%) | -1 | 5/5 | 37 | 0 | 38/38...
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...POTENTIOMETRIC TITRATIONS & SOLUBILITY EQUILIBRIA Tianyi Hu Partner: Yiteng Zhang, Teaching assistant: Nicholas Vizenor Lab section: Thursday 6 pm Date: May 7, 2015 Abstract In this experiment, the burette was used to titrate sodium chloride solution by adding different amount of silver nitrate solution. The electrochemical potential was measured by a reference electrode in order to understand the Nernst equation and calculate out experimental result of concentration of silver nitrate solution, Ksp of Ag ion, and the concentration of chloride ion in the unknown sodium chloride solution. There are four titration performed in this experiment, 2 rough titration of both 150 ppm and unknown sodium chloride solution and 2 careful...
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...Titration Analysis of Commercial Bleach Introduction In order to determine the percentage of sodium hypochlorite in two different commercial bleaches and compare the relative effectiveness of the two bleaches a titration procedure will be used. Titration is a procedure for determining an unknown concentration of a solute using a reaction with a second solution with a known concentration. Commercial bleaches contain sodium hypochlorite. This is the active ingredient. In experiment 7 we demonstrated the ability to do a titration procedure to determine the sodium hypochlorite percentage in two commercial bleaches. Once we figured the two percentages we compared them to see which was the strongest. To determine the amount of NaClO in each solution we had to perform two successive oxidation reduction reactions. For the experiment to have been successful NaClO(aq)+2NaI(aq)+2HC2H3O2(aq)→I2(aq)+NaCl(aq)+2NaC2H3O2(aq)+H2O(l) needed to become colorless shown as: I2(aq)+2Na2S2O3(aq)→Na2S4O6(aq)+2NaI(aq). Before beginning, we thought the second bleach would be stronger. Methods and Materials 1. First, we gathered all of the materials, which included: * Buret Clamp * Buret * Ring Stand * Small Funnel * Beaker for collecting waste materials * 125 mL Erlenmeyer flask 2. Then we rinsed the buret with 10 mL of 0.100 M sodium thiosulfate (Na2S2O3), using the small funnel, to clean the buret. The sodium thiosulfate was discarded into the waste beaker. 3. After...
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...CHEM. 201, EXPERIMENT 4 TITRATION CURVES PROCEDURE: See the pre-lab report on page 15 of my laboratory notebook for an outline of the general procedure. The unknown acid number was 6553, and the concentration of NaOH used in the experiment was .09912 M. Also, three drops of phenolphthalein indicator were added to the initial titration and the titration curve. EXPERIMENTAL DATA: Initial Titration: * Volume of NaOH added at the endpoint was 29.8 mL Titration Curve: * Volume of NaOH added at the endpoint was 29.0 mL CALCULATED RESULTS: Acid concentration from first titration was .118M Ka from initial pH was 1.08x10^-5 Acid concentration from titration curve was .115M Titration | Volume of NaOH (mL) | pH | (base)/(acid) | pKa | Ka | 1/4 | 7.25 | 4.1 | 1/3 | 4.577 | 2.65x10^-5 | 1/2 | 14.5 | 4.6 | 1 | 4.6 | 2.55x10^-5 | 3/4 | 21.8 | 5.19 | 3 | 4.713 | 1.94x10^-5 | Average: | | | | 4.663 | 2.18x10^-15 | DISCUSSION: The purpose of the experiment was to titrate a weak acid of unknown concentration with a strong base, NaOH, and then utilizing an initial titration and titration curve to determine that acid concentration and Ka. After performing the initial titration of the acid concentration, we calculated it to be 0.118 M, with a Ka of 1.08x10^-5. On the other hand, when we performed the titration curve, it calculated an acid concentration of 0.115 M and a Ka of 2.18x10^-5. The results I obtained seemed reasonable...
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...Analysis of Iron Supplement by Redox Titration Purpose: to calculate the percentage of iron in a vitamin supplement by titrating using a KMnO4 solution. Materials: .01 M Potassium Permanganate (KMnO4) Iron Supplement Tablets Distilled water 6 M Sulfuric Acid (H2SO4) 50 mL buret 3 Erlenmeyer Flasks Heat Plate Procedure: First, mass the iron supplement and crush up using a mortar and pestle before adding to 25 mL of distilled water. Next, heat and stir the crushed tablet solution to dissolve it in the water and add 1 mL of 6 M sulfuric acid. Finally, titrate with the 50 mL buret using the potassium permanganate solution and record the volume taken from the buret once neutralized. Conclusion: Titration is the volumetric measurement of a solution of known concentration when it reacts completely with a measured volume or mass of another substance. (Source: dunfried.wikispaces.com). In this experiment, the titrant, the substance doing the titration, is the standardized potassium permanganate solution. This solution will determine the amount of analyte (a chemical substance that is the subject of chemical analysis (source: Merriam-webster.com)) or iron (II) in the supplement pills. Using a balanced redox reaction (one with both an oxidation and reduction of electrons) for this experiment, the percent of iron in the sample can be calculated through basic stoichiometry. This percentage can be compared to the percentage labeled on the bottle of pills to determine...
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...Title Acid-base titration: Determination of the percentages (%) of sodium carbonate (Na2CO3) and sodium hydroxide (NaOH) in a mixture Objective To determine the respective weight per cent of sodium carbonate and sodium hydroxide in a mixture by acid-base titration. Result and calculation Part A Titration 1 Titration number 1 2 3 Initial volume of burette( cm3) 5.10 2.70 9.70 Final volume of burette (cm3) 34.40 31.80 39.20 Total volume of HCl used (cm3) 29.30 29.10 29.50 Average volume of HCl required for titration =(29.30+29.10+29.50)/3 cm3 = 29.30 cm3 Titration 2 Titration number 1 2 3 Initial volume of burette( cm3) 4.50 14.00 2.70 Final volume of burette (cm3) 25.00 21.70 22.80 Total volume of HCl used (cm3) 20.50 20.30 20.10 Average volume of HCl required for titration =(20.50+20.30+20.10)/3 cm3 = 20.30 cm3 Part B Titration number Rough 1 2 3 Initial volume of burette( cm3) 4.9 4.80 3.60 2.20 Final volume of burette (cm3) 28.3 28.90 27.70 26.20 Total volume of HCl used until phenolphthalein decolourised (cm3) , x 23.4 24.10 24.10 24.00 Initial volume of burette after adding methyl orange indicator ( cm3) 28.3 28.90 27.70 26.20 Final volume of burette (cm3) 34.1 33.40 32.10 30.70 Total volume of HCl used until phenolphthalein decolourised (cm3) , y 5.8 4.50 4.40 4.50 Average volume of HCl required to react with Na2CO3 (2y) =2(4.50+4.40+4.50)/3 cm3 = 2(4.4667) cm3 ...
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...Acid-Base Titration: Determination of the Percentages (%) of Sodium Carbonate (Na2Co3) and Sodium Hydroxide (Naoh) in a Mixture Title Acid-base titration: Determination of the percentages (%) of sodium carbonate (Na2CO3) and sodium hydroxide (NaOH) in a mixture Objective To determine the respective weight per cent of sodium carbonate and sodium hydroxide in a mixture by acid-base titration. Result and calculation Part A Titration 1 Titration number 1 2 3 Initial volume of burette( cm3) 5.10 2.70 9.70 Final volume of burette (cm3) 34.40 31.80 39.20 Total volume of HCl used (cm3) 29.30 29.10 29.50 Average volume of HCl required for titration =(29.30+29.10+29.50)/3 cm3 = 29.30 cm3 Titration 2 Titration number 1 2 3 Initial volume of burette( cm3) 4.50 14.00 2.70 Final volume of burette (cm3) 25.00 21.70 22.80 Total volume of HCl used (cm3) 20.50 20.30 20.10 Average volume of HCl required for titration =(20.50+20.30+20.10)/3 cm3 = 20.30 cm3 Part B Titration number Rough 1 2 3 Initial volume of burette( cm3) 4.9 4.80 3.60 2.20 Final volume of burette (cm3) 28.3 28.90 27.70 26.20 Total volume of HCl used until phenolphthalein decolourised (cm3) , x 23.4 24.10 24.10 24.00 Initial volume of burette after adding methyl orange indicator ( cm3) 28.3 28.90 27.70 26.20 Final volume of burette (cm3) 34.1 33.40 32.10 30.70 Total volume of HCl used until phenolphthalein decolourised (cm3) , y 5.8...
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...equivalence point using pH titration of Potassium Hydrogen Phalate and 0.1 N Sodium Hydroxide with phenolphthalein indicator By Lee Binks (12647892), Penny Jinks (12993456) and Hong Links 13504733) Date submitted: Sept 18th , 2013 Abstract- The experiment described elucidates the fundamental principle of potentiometric titrations and determination of equivalence point using pH titration of Potassium Hydrogen Phalate and 0.1 N Sodium Hydroxide with the use of phenolphthalein indicator. The present study revealed the equivalence point by using simple titration curve which was further confirmed by using first and second derivative plots and pKa value of the Potassium Hydrogen Phalate. Introduction- The pH meter measures the pH of a solution and provides a direct method of obtaining a titration curve which is a graph of measured pH values versus the volume of titrant added in milliliters. The equivalence point is the point at which a stoichiometrically equivalent amount of base has been added to the acid. It does not mean that pH will be necessarily 7. KHP is a monoprotic acid. The neutralization with NaOH takes place in a 1:1 ratio HOOCC6H4COOK (aq) + NaOH ( aq) ---> C6H4 ( COO)2 2- (aq) + K+ (aq) + Na+ (aq) The equivalence point occurs in the region where there is a relatively large change in pH with a relatively small change in volume on the titration curve. The steeper the curve the more precisely equivalence point can be established. Once a titration curve is constructed...
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