Question

DULLOH HILL UCUREI LULUS solution in the beaker. 13. When you have finished the testing, dispose of the reaction mixture as directed. Rinse the Sensor with distilled water in preparation for a second trial. Repeat the necessary steps to CS a new sample of the acetic acid solution. DATA TABLE Titration Results Trial 1 Equivalence point pH Trial 2 9.69.16.24 15.675.53 pH of half-titrated solution DATA ANALYSIS 1. Calculate the pKa and K, using the results of your testing. 2. Find the accepted values for the pKa and Ka of acetic acid. How well do the accepted values compare with your calculated values? Explain. 3. Explain why the pH at the half-titration point is equal to the pk, in your experiment. 4. Explain how this test could be done using only an indicator solution and no electronic means of measuring pH.

Answer 1

Solution:

1) In the titration of a weak acid, the pKa of the acid is equivalent with the pH at half equivalence point.

Thus,

>>>pKa for trial 1 = 5.67

Hence, Ka = 10 ^-pka = 10^-5.57

Ka = 2.14 x 10^-6

>>>pKa for trial 2 = 5.53

Hence, Ka = 10 ^-pka = 10^-5.53

Ka = 2.95 x 10^-6

2) The accepted value of Ka and pka fior acetic acid are:

pKa = 4.74

Ka = 1.8 x 10^-5

3) At half equivalence point, the concenmtrations of CH3COOH and its anion CH3COO- is equal.

Hence, according to Hendersen- Haseelbalch equation,

pH = pKa + log [ CH3COO-] / [CH3COOH]

pH = pka + 0

pH = pKa

4) Indicators are the chemical agents which changes its color even in slight change in pH. Since, acetic acid is a weak acid and it forms buffer on its dissociation. This will causes a slight change in pH on addition of base or acid. Therefore, indicator is more suitable for detection by changing its color.