Chemistry
jessiereid541
2016-04-10 16:22:27
The ka value for acetic acid, ch3cooh(aq), is 1.8Ã 10â5 m. calculate the ph of a 1.40 m acetic acid solution.
ANSWERS
revoltingriot
2016-04-10 18:16:02

1) Equlibrium reaction CH3COOH (aq) = CH3COO(-) (aq) + H(+) (aq) 2) Equilibrium constant Keq = Ka = [CH3COO-] [H+] / [CH3COOH] 3) Equilibrium concentrations                   CH3COOH      CH3COO-     H+ start               1.40                  0              0 react                  x                   0              0 produced           0                   x              x equilibrium     1.40 - x            x               x => Ka = x * x / (1.40 - x) Approximation: given that Ka is very small x <<< 1,40 and 1.40 - x  ≈ 1.40 => Ka ≈ x^2 / 1.40 => x^2 ≈ 1.40Ka = 1.40 * 1.8 * 10^ - 5 = 2.52 * 10^-5 => x ≈ √(2.52 * 10^-5) ≈ 5.02 * 10^ -3 M 4) pH = log 1 / [H+] [H+] = x = 5.02 * 10^-3M => pH ≈ log (1 / 5.02 * 10^-3) ≈ 2.3 Answer: 2.3

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