Question

# A beaker with 2.00×102 mL of an acetic acid buffer with a pH of 5.000 is...

A beaker with 2.00×102 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 6.60 mL of a 0.400 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

In the initial buffer

[CH3COOH] + [CH3COO-] = 0.100 mol/l and their quantity is

0.100 mol/L x 0.200 L = 0.0200 mol

Calculate the ratio [CH3COO-]/[CH3COOH] using the buffer formula:

log ([CH3COO-]/[CH3COOH]) = pH – pKa = 5.000 – 4.740 = 0.260

[CH3COO-]/[CH3COOH] = 100.260= 1.820 = 1.820:1

0.0200 mol x 1.820/(1.820 + 1) = 0.0129 mol CH3COO-

0.0200 mol x 1 /(1.820 + 1) = 0.0071 mol CH3COOH

In the initial buffer.

The quantity of HCl added is

0.00660 L x 0.400 mol/L = 0.00264 mol

After neutralization, the buffer composition is:

0.0129 mol - 0.00264 = 0.01026 M CH3COO-

0.0071 mol + 0.00264 = 0.00974 M CH3COOH

The new pH is

pH = pKa + log([CH3COO-]/[CH3COOH])

= 4.740 + log(0.01026 M/0.00974 M)=

= 4.740 + 0.0226 = 4.763

The pH change is 4.763 – 5.000 = - 0.237 pH units.