Question

A 1.41 L buffer solution consists of 0.253 M propanoic acid and 0.110 M sodium propanoate. Calculate the pH of the solution following the addition of 0.061 mol HCl . Assume that any contribution of the HCl to the volume of the solution is negligible. The K a of propanoic acid is 1.34 × 10 − 5 .

Answer 1

pH = 5.01

Step-by-step explanation:

The reaction between the sodium propanoate and the HCl added is the following:

CH₃CH₂COO⁻ + H₃O⁺ ⇄ CH₃CH₂COOH + H₂O

initial 0.110M 0.061moles 0.253M

The number of moles of the acid propanoic (a) and sodium propanoate (b) is:

`\eta_(a) = [a]*Va = 0.110 M * 1.41 L = 0.155 moles \thinspace acid`

`\eta_(b) = [b]*Vb = 0.253 M * 1.41 L = 0.357 moles\thinspace propanoate`

After the adding of HCl, the number of moles of acid propanoic and propanoate is:

`\eta_(b) = 0.357 moles - 0.061 moles = 0.296 moles \thinspace propanoate`

`\eta_(a) = 0.155 moles + 0.061 moles = 0.216 moles \thinspace acid`

Hence, the pH of the solution after the addition of HCl is:

`pH = pKa + log((b)/(a)) = -log(1.34 \cdot 10^(-5)) + log((0.296)/(0.216)) = 5.01`

Therefore, the pH of the solution is 5.01.

I hope it helps you!