Question

A beaker with 115 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 7.60 mL of a 0.440 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

pH = 5.000

pKa = 4.740

CH3COOH + CH3COO- = 115 x 0.1 = 11.5

pH = pKa + log [salt / acid]

5 = 4.740 + log [CH3COO- / CH3COOH]

[CH3COO- / CH3COOH] = 1.8197

CH3COOH + CH3COO- = 11.5

CH3COOH + 1.8197 CH3COOH = 11.5

millimoles of [CH3COOH] = 4.078

millimoles of [CH3COO-] = 7.422

millimoles of HCl = 7.60 x 0.440 = 3.344

pH = pKa + log [salt - C / acid + C]

    = 4.74 + log [7.422 - 3.344 / 4.078 + 3.344]

pH = 4.480

pH change = 4.480 - 5 = - 0.52

pH change = - 0.52