pH of a buffer help?

2016-10-23 12:53 am

A 1.46 L buffer solution consists of 0.342 M propanoic acid and 0.103 M sodium propanoate. Calculate the pH of the solution following the addition of 0.071 moles of HCl. Assume that any contribution of the HCl to the volume of the solution is negligible. The Ka of propanoic acid is 1.34 × 10-5.
I got 2.54 but that's incorrect.

回答 (1)

胡雪8°

2016-10-23 3:05 pm

✔ 最佳答案

In the 1.46 L buffer solution :
[CH₃COOH] = 0.342 M
[CH₃COO⁻] = 0.103 M

HCl is a strong acid which completely ionizes in water to give hydrogen ions.
[H⁺] added to the buffer solution = (0.071 mol) / (1.46 mol/L) = 0.049 M

Equation of the reaction :
CH₃COO⁻(aq) + H⁺(aq) ⇌ CH₃COOH(aq)
After reaction :
[CH₃COOH] = (0.342 + 0.049) M = 0.390 M
[CH₃COO⁻] = (0.103 - 0.049) M = 0.054 M

Consider the dissociation of CH₃COOH.
CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) …. Kₐ

pH = pKₐ - log([CH₃COOH]/[CH₃COO⁻]) = -log(1.34 × 10⁻⁵) - log(0.390/0.054) = 4.01

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