Question

Calculate the volume of 3 M HCl needed to change the pH of 75 mL of the undiluted buffer solution by one pH unit (buffer capacity).

[undiluted buffer solution is prepared by mixing 100 mL of 1.060 M acetic acid with 100 mL of 1.068 M sodium acetate solution]

Answer 1

initial pH

pH = pKa + log(acetate/acid)

pH = 4.75 + log(1.068/1.060) = 4.75326538743

if we want to decrease by 1 unit

pH new = pH old - 1 = 4.75326538743-1 = 3.7532

then

pH = 4.75 + log(acetate/acid)

3.7532 = 4.75 + log(acetate/acid)

solve for ratio

acetate/acid = 10^(3.7532-4.75) = 0.1007

acetate = 0.1007*acid

and

mmol of acetate initially = MV = 75*1.068 = 80.1

mmol of acid initially = MV = 75*1.060 = 79.5

after adding

mmol of acid = MV = 3*V

then

mmol of acetate = 80.1 - 3*V

mmol of acid = 79.5 + 3V

then

from ratio

acetate = 0.1007*acid

80.1 - 3*V = (0.1007)(79.5 + 3V)

80.1-8.00565 = 3V -0.3021V

72.09 = 2.6979V

V = (72.09/2.6979) = 26.720 ml of acid must be added