Question

calculate the pH of a solution of 0.25 M acetic acid and 0.34 M sodium acetate before and after 0.036 M NaOH is added to the solution. Ka of HAc=1.8*10^-5

Answer 1

Before addition of NaOH:

pH = pKa + log [ sodium acetate]/[acetic acid]

= -log Ka +  log [ sodium acetate]/[acetic acid]

= - log (1.8*10^-5) + log [0.34]/[0.25]

= 4.87

pH = 4.87

After addition of 0.036 M NaOH:

CH3COOH + NaOH ------------------> CH3COONa + H2O

Hence,

pH = pKa + log [ sodium acetate] + [NaOH] /[acetic acid] - [NaOH]

= -log Ka +  log [ sodium acetate] + [NaOH] /[acetic acid] - [NaOH]

= - log (1.8*10^-5) + log { [0.34] + [0.036] /[0.25] -[0.036]}

= 4.98

pH = 4.98