Chemistry
ahamtan36
2016-04-09 19:25:12
What is the ph of a solution of 0.50 m acetic acid?
ANSWERS
Gemzy2002
2016-04-10 01:26:34

You need to use the Ka for the acetic acid and the equilibrium equation. Ka = 1.85 * 10^ -5 Equilibrium reaction: CH3COOH (aq) ---> CH3COO(-) + H(+) Ka = [CH3COO-][H+] / [CH3COOH] Molar concentrations at equilibrium CH3COOH         CH3COO-     H+  0.50  - x                  x                 x Ka = x*x / (0.50 - x) = x^2 / (0.50 - x) Given that Ka is << 1 => 0.50 >> x and 0.50 - x ≈ 0.50 => Ka ≈ x^2 / 0.50 => x^2 ≈ 0.50 * Ka = 0.50 * 1.85 * 10^ -5 = 0.925 * 10^ - 5 = 9.25 * 10 ^ - 6 => x = √ [9.25 * 10^ -6] = 3.04 * 10^ -3 ≈ 0.0030 pH = - log [H+] = - log (x) = - log (0.0030) = 2.5 Answer: 2.5

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