Question

A 1.37 L buffer solution consists of 0.103 M butanoic acid and 0.349 M sodium butanoate. Calculate the pH of the solution following the addition of 0.080 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The Ka of butanoic acid is 1.52 × 10^-5.

Answer 1

Ka of butanoic acid is 1.52 × 10^-5.

pKa = - log Ka = - log 1.52 × 10^-5.

= 4.82

Moles of butanoic acid = 0.103 M x 1.37 L

= 0.1411

Moles of butanoate = 0.349 x 1.37 L

= 0.4781

Moles OH- added = 0.080

                   HA    +      OH-     ==> A- +    H2O

After reaction:

Moles of butanoic acid = 0.1411 - 0.080 = 0.0611

Moles of butanoate = 0.4781 + 0.080 = 0.5581

Total volume = 1.37 L

[HA] = 0.0611/ 1.37 = 0.0445 M

[A-] = 0.5581 / 1.37 = 0.4073 M

pH = pKa + log(base/acid)

pH = 4.82 + log 0.4073 / 0.0445

pH = 4.82 + log 9.15

pH = 4.82 + 0.961

= 5.781