If you add 250 mL of 0150 M HCL to 100 mL of a buffer consis

If you add 2.50 mL of 0.150 M HCL to 100 mL of a buffer consisting of 0.100 M CH₃COOH and 0.200 M CH₃COONa, what will be the change in pH? Ka of CH₃COOH is 1.8e-5.

Solution

Buffer solution

CH3COOH/CH3COONa = acid/base

[CH3COOH] = 0.100 M

[CH3COONa] = 0.200 M

pKa = -log(Ka) = -log(1.8 x 10^-5) = 4.74

Using Hendersen-Hasselbalck equation,

initial pH of buffer solution

pH = pKa + log(base/acid)

pH = 4.74 + 0.200/0.100 = 5.04

After HCl = 0.150 M x 2.50 ml = 0.375 mmol was added

[CH3COONa] = 0.200 M x 100 ml - 0.375 = 19.625 mmol

[CH3COOH] = 0.100 M x 100 ml + 0.375 = 10.375 mmol

new pH,

pH = 4.74 + log(19.625/10.375) = 5.02

Change in pH = 5.04 - 5.02 = 0.02