Question

A buffer is formed by mixing 450 mL of a 0.388 M sodium formate solution with 375 mL of a 0.450 M formic acid, HCHO2, solution. What is the pH of the new solution?
Correct Answer: 3.78
-I just want to know the process for the problem. Thanks!

Answer 1

Molarity = Moles / Liter

Moles = Molarity x Liter

450 mL of a 0.388 M sodium formate

Moles of sodium formate = 0.388 M x (450/1000 L) = 0.1746 moles

375 mL of a 0.450 M formic acid

Moles of formic acid = 0.450 M x (375/1000 L) = 0.16875 moles

Total volume = 450 mL + 375 mL = 825 mL = 0.825 L

So, the concentration terms are

[sodium formate] = 0.1746 moles / 0.825 L = 0.2116 M

[formic acid] = 0.16875 moles / 0.825 L = 0.2045 M

Using Henderson-Hesselbalach equation

pH = pKa + log { [salt] / [acid] }

pH = pKa + log { [sodium formate] / [formic acid] }

Formic acid has a pKa of 3.75

So,

pH = pKa + log { [sodium formate] / [formic acid] }

pH = 3.75 + log { 0.2116 M / 0.2045 M }

pH = 3.75 + log { 1.035 }

pH = 3.75 + 0.02

pH = 3.77