The Ka of propanoic acid (C2H5COOH) is 1.34 × 10-5. Calculate the pH of the solution and the concentrations of C2H5COOH and C2H5COO– in a 0.421 M propanoic acid solution at equilibrium.
construct ICE table
C2H5COOH (aq) + H2O (l) <------------> C2H5COO- (aq) + H3O+ (aq)
I 0.421 0 0
C -x +x +x
E 0.421 - x +x +x
Ka = [C2H5COO-] [H3O+] / [C2H5COOH]
1.34 x 10^-5 = [x] [x] / [0.421-x]
x2 + x 1.34 * 10^-5 - 5.6414 * 10^-6 = 0
solve the quadratic equation
x = 0.002368 M
so concentrations of
[C2H5COO-] = x = 0.002368 M
[H3O+] = x = 0.002368 M
[C2H5COOH] = 0.421 - 0.002368 = 0.4186 M
pH = -log[H3O+]
= -log[0.002368]
= 2.625
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