Question

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 0.551 M propanoic acid solution at equilibrium.

Answer 1

C2H5COOH ---> C2H5COO-   +   H+
0.551                          0                        0   (initial)
0.551-x                      x                        x    (at equilibrium)

Ka = [C2H5COO-][H+] / [C2H5COOH ]
1.34*10^-5 = x*x / (0.551-x)

Here Ka is very small, so x will be very small and it can be ignored as compared to 0.551
So, above expression becomes
1.34*10^-5 = x*x / 0.551
x=2.72*10^-3 M

Answer:
[C2H5COO–] = x = 2.72*10^-3 M
[C2H5COOH ] = 0.551-x= 0.551-2.72*10^-3 = 0.548 M