Question

A beaker with 145 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 4.50 mL of a 0.350 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760. ΔpH =

Answer 1

Initial pH = 5.00

Initial concentrations of [CH3COO-] and [CH3COOH]

pH = pKa + log(base/acid)

5.00 = 4.760 + log([CH3COO-]/[CH3COOH])

[CH3COO-] = 1.74[CH3COOH]

[CH3COOH] + [CH3COO-] = 0.1 M x 145 ml = 14.5 mmol

[CH3COOH] + 1.74[CH3COOH] = 14.5 mmol

[CH3COOH] = 5.292 mmol

[CH3COO-] = 9.208 mmol

When HCl added = 0.350 M x 4.50 ml = 1.575 mmol

New [CH3COOH] = (5.292 + 1.575) mmol/(145 + 4.5) ml = 0.046 M

New [CH3COO-] = (9.208 - 1.575) mmol/(145 + 4.5) ml = 0.051 M

New pH = 4.760 + log(0.051/0.046)

              = 4.80

Change in pH = initial pH - final pH = 5.00 - 4.80 = 0.20

[pl. note If we to consider change in pH = (final - initial)pH = -0.20]