Question

Calculate the pH of 1.00 L of the buffer 0.95 M CH3COONa/0.92 M CH3COOH before and after the addition of the following species. (Assume there is no change in volume.) (a) pH of starting buffer: (b) pH after addition of 0.040 mol NaOH: (c) pH after further addition of 0.100 mol HCl:

Answer 1

Answer: (a) pH = 4.774, (b) pH = 4.811 and (c) pH = 4.681

Explanation: (a) pH of the buffer solution is calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

pKa for acetic acid is 4.76. concentration of base and acid are given as 0.95M and 0.92M. Let's plug in the values in the equation and calculate the pH of starting buffer.

$pH=4.76+log(\frac{0.95}{0.92})$

pH = 4.76 + 0.014

pH = 4.774

(b) When 0.040 mol of NaOH (strong base) are added to the buffer then it reacts with 0.040 mol of acetic acid and form 0.040 mol of sodium acetate.

Original buffer volume is 1.00 L. So, the original moles of sodium acetate will be 0.95 and acetic acid will be 0.92.

moles of acetic acid after addition of NaOH = 0.92 - 0.040 = 0.88

moles of sodium acetate after addition of NaOH = 0.95 + 0.040 = 0.99

Let's again plug in the values in the Handerson equation:

$pH=4.76+log(\frac{0.99}{0.88})$

pH = 4.76 + 0.051

pH = 4.811

(c) When 0.100 mol of HCl are added then it reacts with exactly 0.100 moles of sodium acetate(base) and form 0.100 moles of acetic acid(acid).

so, new moles of acetic acid = 0.92 + 0.100 = 1.02

new moles of sodium acetate = 0.95 - 0.100 = 0.85

Let's plug in the values in the equation:

$pH=4.76+log(\frac{0.85}{1.02})$

pH = 4.76 - 0.079

pH = 4.681