Question

The K_a of propanoic acid (C₂H₅COOH) is 1.34 x 10^-5. Calculate the pH of the solution and the concentrations of C₂H₅COOH and C₂H₅COO^- in a 0.297 M propanoic acid solution at equilibrium.

pH = _____

[C₂H₅COOH] = _____ M

[C₂H₅COO^-] = ______ M

Answer 1

..........................[C2H5COOH] ............. [H+] ........... [C2H5C00-]
initial .......................... 0.227 ....... ........... 0............ 0
change ... ................ -x .......... ........ ......... +x .... ...... .... + x
equilibrium......... 0.297 -x ........ ................. x ..... .......... x

Ka = 1.34 x 10^-5 = [H+][C2H5COO-]/[C2H5COOH]

1.34 x 10^-5 = (x)(x)/(0.297-x)

As an approximation, I am neglecting x in the denominator rather than solving a quadratic.

1.34 x 10^-5 = x^2/0.297

3.9798 x 10^-6 = x^2

x = 1.9949 x 10^-3

[H+] = 1.9949 x 10^-3

pH = - log[H+] = - log(1.9949 x 10^-3)

pH = 2.702