Question

1. How will the equivalence point volume change if you titrate the two solutions in question 2? What is the pH of the equivalence point of the two solutions if you titrate with 0.3 M NaOH?

Note: You do not need to answer question 2 in order to do question 3.
The two solutions that question #3 refers to are:
(1) a 10-mL vinegar solution that has a concentration of 5% (w/v%)
(2) a 10-mL vinegar solution that has a concentration of 5% (w/v%) together with 30mL of water

Each one of these solutions is then titrated separately with the NaOH solution. Ka for acetic acid is listed in question #1 ii. Calculate the amount (in mL) of a 1.520M NaOH that is required to add the following acetic acid solutions to prepare a buffer with the corresponding pH: pKa of acetic acid = 4.74

1. 30.00mL of a 5.00% (w/v%) acetic acid; the resulting acetate buffer has a pH of 5.75
2. 50.00mL of a 5.00% (w/v%) acetic acid; the resulting acetate buffer has a pH of 4.98
3. 40.00mL of a 5.00% (w/v%) acetic acid; the resulting acetate buffer has a pH of 4.33

Hint: Set up an equilibrium table for a reaction between acetic acid and NaOH (similar to the examples that we did in lectures). However, in this case, the mole of NaOH in the equilibrium table is a function of volume (i.e. MNaOH.V) and it will be the limiting reagent in the equilibrium table. You will then use the Henderson-Hasselbalch equation to determine the volume of NaOH (i.e. V) that is required to prepare the specific acetate buffer solution at the corresponding pH.

Answer 1

1. In case of 10 ml of 5% (w/v) vinegar solution the pH observed at equivalence point would be higher (more basic) than the equivalence point (less basic) for the titration of 10 ml of 5% (w/v) vinegar with 30 ml water.

pH at equivalence point

(1) molar concentration of acetic acid solution = 0.5/0.01 x 60.05 = 0.833 M

Volume of NaOH used = 0.833 x 10/0.3 = 0.028 ml

[CH3COONa] = 0.833/0.038 = 22.06 M

1.75 x 10^-5 = x^2/22.06

x = [OH-] = 0.0196 M

pOH = 1.71

pH = 14 - pOH = 12.30

(2) molar concentration of acetic acid solution = 0.5/0.01 x 60.05 = 0.833 M

final concentration of acetic acid solution = 0.833 x 10/40 = 0.21 M

Volume of NaOH used = 0.21 x 40/0.3 = 0.028 ml

[CH3COONa] = 0.21/0.038 = 5.53 M

1.75 x 10^-5 = x^2/5.53

x = [OH-] = 9.83 x 10^-3 M

pOH = 2.01

pH = 11.99

Volume of NaOH required to add to,

1. [CH3COOH] = 1.5/0.03 x 60.05 = 0.833 M

let x amount of base is added

pH = pKa + log([base]/[acid])

5.75 = 4.74 + log(1.520x/0.833 x 30)

255.72 = 1.520x

x = 168.24 ml

Volume of NaOH required = 168.24 ml

2. [CH3COOH] = 2.5/0.05 x 60.05 = 0.833 M

let x amount of base is added

pH = pKa + log([base]/[acid])

4.98 = 4.74 + log(1.520x/0.833 x 50)

72.38 = 1.520x

x = 47.62 ml

Volume of NaOH required = 47.62 ml

3. [CH3COOH] = 2.0/0.04 x 60.05 = 0.833 M

let x amount of base is added

pH = pKa + log([base]/[acid])

4.33 = 4.74 + log(1.520x/0.833 x 40)

12.963 = 1.520x

x = 8.53 ml

Volume of NaOH required = 8.53 ml