Question

A 1.00 L buffer solution is 0.112 M in acetic acid and 0.112 M in sodium acetate. Acetic acid has a pKa of 4.74. What is the pH change of this solution upon addition of 0.1 mol of HCl?

Answer 1

ΔpH = 1.25

Step-by-step explanation:

Using Henderson-Hasselbalch formula:

pH = pka + log [CH₃COONa] / [CH₃COOH]

pH = 4.74 + log [0.112] / [0.112]

pH = 4.74

The reaction of sodium acetate (CH₃COONa) with HCl is:

CH₃COONa + HCl → CH₃COOH + NaCl

Producing acetic acid, CH₃COOH.

If 0.1mol of HCl reacts the final moles of CH₃COONa are:

0.112mol - 0,1 mol = 0.012mol

Moles of acetic acid are:

0.112mol + 0,1 mol = 0.212mol

Using Henderson-Hasselbalch formula:

pH = pka + log [CH₃COONa] / [CH₃COOH]

pH = 4.74 + log [0.012] / [0.212]

pH = 3.49

Change in pH, ΔpH = 4.74 - 3.49 = 1.25

I hope it helps!