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Determination of Equivalence Point Using Ph Titration of Potassium Hydrogen Phalate and 0.1 N Sodium Hydroxide with Phenolphthalein Indicator

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Analytical Chemistry Laboratory I Fall Semester University of Massachusetts Lowell Department of Chemistry

Determination of 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 and the equivalence point is established , an indicator can be used to give a suitable endpoint ( point at which the indicator changes colour ).

Source : http://www.titrations.info/img/HCl-NaOH-01-phenolphthalein.png
In this experiment , we obtained a titration curve of measured pH against the volume of
0.1N NaOH solution added. The equivalence point is established using the steepest tangent to the smooth curve where the pH changes rapidly. The equivalence point is the mid point between the two lines intersecting the volume axis. The equivalence point selected using this method is a more accurate method than using an indicator in the titration. Additionally , pKa values can also be directly read from the titration curve.
A second method used to determine the equivalence point is by using the first derivative method.
To use this method, a graph is constructed of Delta pH / Delta V vs Average volume. The volume at the point where the graph reaches the maxima is the equivalence point of the titration.
In the second derivative of the titration curve, zero crossing is the equivalent point.
Source : http://www2.volstate.edu/chem/1120/Titration-7.gif pH meter consists of three parts- pH electrode ( Reference and Indicator electrode )
,Thermoprobe and Voltmeter. Reference electrode is usually made up of Ag or AgCl and has a fixed potential value while the Indicator Electrode is made up of glass and the potential changes with the concentration of protons . Thermoprobe is used to measure the temperature of the solution and the Voltmeter is used to measure the potential difference between the two electrodes.
Experiment
1.5152 g of dried KHP ( Fisher Scientific Co, Pittsburgh,PA ) was weighed and transferred to a
250 mL volumetric flask to prepare a stock solution in water. 50.0 mL of the KHP stock solution is pipetted into a 250 mL beaker with a magnetic stirring bar and 50 mL of distilled water was added to it. Solution is titrated with 0.1005 N NaOH ( Fisher Scientific Co, Pittsburgh,PA ), in 2 mL base aliquots for the first 10 ml and then in 0.5 mL base aliquots for the next 5 ml as end- point approached at 15 ml. Volume and pH after each addition was recorded by using the calibrated pH meter ( Mettler Toledo , Ohio ). Base was added drop wise until pink end point was observed . Addition of base was continued till five more 1 mL aliquots were added and the volume and pH was recorded after each addition.

Results
TABLE 1 Sample Titration Data Volume of NaOH ( ml ) | pH | 0.0 | 4.03 | 2.0 | 4.63 | 4.0 | 4.86 | 6.0 | 5.09 | 8.0 | 5.30 | 10.0 | 5.53 | 10.5 | 5.61 | 11.011.512.012.513.013.514.014.515.015.516.016.517.018.019.020.021.0 | 5.685.765.845.976.096.266.567.109.9510.5110.7410.9010.9911.1611.2711.3611.43 |

Table 2 Titration Curve between 0.1 N NaOH and pH

Calculation :
At half Equivalence point,
[HA] = [A-]
[H3O + ] = Ka
Therefore, pH= pKa
Equivalence Volume = 15 ml
Therefore Half Equivalence Volume = 7.5 ml
V = 6.0, pH = 5.09
V = 8.0, pH = 5.30 pH of the Half Equivalence Volume = 5.2 = pKa
Therefore, Ka = 6.30 x 10 ^ - 6
Tabls 3 First Derivative Table- Vol in ml | pH | Delta pH | Delta V | Mean Volume in ml | Delta pH / Delta V | 0 | 4.03 | | | | | | | 0.6 | 2 | 1 | 0.3 | 2 | 4.63 | | | | | | | 0.23 | 2 | 3 | 0.115 | 4 | 4.86 | | | | | | | 0.23 | 2 | 5 | 0.115 | 6 | 5.09 | | | | | | | 0.21 | 2 | 7 | 0.105 | 8 | 5.3 | | | | | | | 0.23 | 2 | 9 | 0.115 | 10 | 5.53 | | | | | | | 0.08 | 0.5 | 10.25 | 0.16 | 10.5 | 5.61 | | | | | | | 0.07 | 0.5 | 10.75 | 0.14 | 11 | 5.68 | | | | | | | 0.08 | 0.5 | 11.25 | 0.16 | 11.5 | 5.76 | | | | | | | 0.08 | 0.5 | 11.75 | 0.16 | 12 | 5.84 | | | | | | | 0.13 | 0.5 | 12.25 | 0.26 | 12.5 | 5.97 | | | | | | | 0.12 | 0.5 | 12.75 | 0.24 | 13 | 6.09 | | | | | | | 0.17 | 0.5 | 13.25 | 0.34 | 13.5 | 6.26 | | | | | | | 0.3 | 0.5 | 13.75 | 0.6 | 14 | 6.56 | | | | | | | 0.54 | 0.5 | 14.25 | 1.08 | 14.5 | 7.1 | | | | | | | 2.85 | 0.5 | 14.75 | 5.7 | 15 | 9.95 | | | | | | | 0.56 | 0.5 | 15.25 | 1.12 | 15.5 | 10.51 | | | | | | | 0.23 | 0.5 | 15.75 | 0.46 | 16 | 10.74 | | | | | | | 0.16 | 0.5 | 16.25 | 0.32 | 16.5 | 10.9 | | | | | | | 0.09 | 0.5 | 16.75 | 0.18 | 17 | 10.99 | | | | | | | 0.17 | 1 | 17.5 | 0.17 | 18 | 11.16 | | | | | | | 0.11 | 1 | 18.5 | 0.11 | 19 | 11.27 | | | | | | | 0.09 | 1 | 19.5 | 0.09 | 20 | 11.36 | | | | | | | 0.07 | 1 | 20.5 | 0.07 | 21 | 11.43 | | | | | | | | | | |
Table 4 First Derivative Plot-

Table 5 Second Derivative Table- Mean Vol ml | Delta pH/Delta V | Delta (Delta V) | Delta (Delta pH/Delta V ) | V Ave | Delta (Delta pH/Delta V )/ Delta ( Delta V ) | | | | | | | 1 | 0.3 | | | | | | | 2 | 0.185 | 2 | 0.0925 | 3 | 0.115 | | | | | | | 2 | 0 | 4 | 0 | 5 | 0.115 | | | | | | | 2 | 0.01 | 6 | 0.005 | 7 | 0.105 | | | | | | | 2 | -0.01 | 8 | -0.005 | 9 | 0.115 | | | | | | | 1.25 | -0.05 | 9.63 | -0.04 | 10.25 | 0.16 | | | | | | | 0.5 | 0.02 | 10.5 | 0.04 | 10.75 | 0.14 | | | | | | | 0.5 | -0.02 | 11 | -0.04 | 11.25 | 0.16 | | | | | | | 0.5 | 0 | 11.5 | 0 | 11.75 | 0.16 | | | | | | | 0.5 | -0.1 | 12 | -0.2 | 12.25 | 0.26 | | | | | | | 0.5 | 0.02 | 12.5 | 0.04 | 12.75 | 0.24 | | | | | | | 0.5 | -0.1 | 13 | -0.2 | 13.25 | 0.34 | | | | | | | 0.5 | -0.26 | 13.5 | -0.52 | 13.75 | 0.6 | | | | | | | 0.5 | -0.48 | 14 | -0.96 | 14.25 | 1.08 | | | | | | | 0.5 | -4.62 | 14.5 | -9.24 | 14.75 | 5.7 | | | | | | | 0.5 | 4.58 | 15 | 9.16 | 15.25 | 1.12 | | | | | | | 0.5 | 0.66 | 15.5 | 1.32 | 15.75 | 0.46 | | | | | | | 0.5 | 0.14 | 16 | 0.28 | 16.25 | 0.32 | | | | | | | 0.5 | 0.14 | 16.5 | 0.28 | 16.75 | 0.18 | | | | | | | 0.75 | 0.01 | 17.13 | 0.013 | 17.5 | 0.17 | | | | | | | 1 | 0.06 | 18 | 0.06 | 18.5 | 0.11 | | | | | | | 1 | 0.02 | 19 | 0.02 | 19.5 | 0.09 | | | | | | | 1 | 0.02 | 20 | 0.02 | 20.5 | 0.07 | | | | | | | | | | | |

Table 6 Second Derivative Plot-

Discussion
At the equivalence point, the slope of the titration curve is at a maximum and the rate of change of pH with addition of titrant is at its highest .So, we plotted the rate of change of pH with change in volume (ΔpH/ΔV) against volume to get the first derivative spiked curve whose peak occur at the equivalence point. The equivalence point is the maximum which occurs at 15 ml. Plotting first derivative graph however leads to an uncertainty that can be partially avoided by a extrapolating another graph known as second derivative plot .
The second derivative of a titration curve is the rate of change of the first derivative with respect to the change in the average volume and pass through zero at the equivalence point which is 15 ml. The average of the two successive volumes used for the first derivative plot is also used for the second derivative plot (Table 5). In using derivative methods, the volume increment should not be too large . There should be sufficient points near the equivalence point. Also using derivatives results in the increase in noise esp the second derivative.
References
1. "Quantitative Chemical Analysis" by Daniel C. Harris, 6th Edition, 2003, W. H. Freeman and Co. 2. Laboratory Manual

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