Question

Calculate the pH of a solution of 0.25 M Acetic acid and 0.34M Sodium acetate before and after 0.036 M NaOH is added to the solution. Ka of HAc = 1.8 x 10 -5

Please Use the ICE method

Answer 1

Before adding NaOH

                        CH3COOH     +    H2O H₃O⁺    +    CH₃COO^-

Initial                    0.25                                           0                0.34

Change               0.25-x                                          +x            0.34 +x

Equilibrium           0.25-x                                           x             0.34+x

Ka = 1.8 x 10^-5 = [CH₃COO^-] [ H₃O⁺] / [CH3COOH]   = (0.34+x)(x)/(0.25-x) (0.34)(x) / (0.25)

We have assumed that x is small compared with the initial concentration of acid and conjugate base as Ka is very small.

x = [H₃O⁺] = Ka [CH3COOH] / [CH₃COO^-] = (1.8 x 10^-5) x 0.25 / 0.34 = 1.32 x 10^-5 M

pH = -log [H₃O⁺] = -log(1.32 x 10^-5) = -log1.32 + 5 = 4.88

After adding NaOH

Added NaOH is a strong base that will convert acetic acid to sodium acetate, so after the addtion of NaOH concentration will be given as follow:

[CH3COOH] = 0.25 - 0.036 = 0.214 M

[CH₃COO^-] = 0.34 + 0.036 = 0.376 M

x = [H₃O⁺] = Ka [CH3COOH] / [CH₃COO^-] = (1.8 x 10^-5) x 0.214 / 0.376 = 1.02 x 10^-5 M

pH = -log [H₃O⁺] = -log(1.02 x 10^-5) = -log1.02 + 5 = 4.99