Calculating pH of acid?

Question

Given the Ka value for ethanoic acid (CH3C00H) to be 1.8 x 10^-5 calculate the concentration at equilibrium of each constituent in a 0.1M solution of the acid. Then calculate the pH of the acid?

Answer 1

CH3COOH <----> CH3COO- + H+

initial concentration

0.1

change

-x .. . . . . . . . . .. . . . . . . . +x .. . . . .+x

at equilibrium

0.1-x. . . . . . . ..   .. . . . . . ..x. . . . . . . x

Ka = 1.8 x 10^-5 = (x)(x) / 0.1-x

x = [H+] = [CH3COO-] =0.00134 M

pH = - log 0.00134 =2.87

[CH3COOH] = 0.1 - 0.00134 =0.0987 M

Answer 2

Actually

It is easy

CH3COOH <----> CH3COO- + H+

initial concentration

0.1

change

-x .. . . . . . . . . .. . . . . . . . +x .. . . . .+x

at equilibrium

0.1-x. . . . . . . .. .. . . . . . ..x. . . . . . . x

Ka = 1.8 x 10^-5 = (x)(x) / 0.1-x

x = [H+] = [CH3COO-] =0.00134 M

pH = - log 0.00134 =2.87

[CH3COOH] = 0.1 - 0.00134 =0.0987 M