The ka value for acetic acid, ch3cooh(aq), is 1.8ã 10â5 m. calculate the ph of a 1.40 m acetic acid solution.

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

2.3

Ithink the answer is 56.

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