Question

A 1.49 L buffer solution consists of 0.186 M butanoic acid and 0.258 M sodium butanoate. Calculate the pH of the solution following the addition of 0.062 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

mol of NaOH added = 0.062 mol

C3H8COOH will react with OH- to form C3H8COO-

Before Reaction:

mol of C3H8COO- = 0.258 M *1.49 L

mol of C3H8COO- = 0.3844 mol

mol of C3H8COOH = 0.186 M *1.49 L

mol of C3H8COOH = 0.2771 mol

after reaction,

mol of C3H8COO- = mol present initially + mol added

mol of C3H8COO- = (0.3844 + 0.062) mol

mol of C3H8COO- = 0.4464 mol

mol of C3H8COOH = mol present initially - mol added

mol of C3H8COOH = (0.2771 - 0.062) mol

mol of C3H8COOH = 0.2151 mol

Ka = 1.52*10^-5

pKa = - log (Ka)

= - log(1.52*10^-5)

= 4.8182

since volume is both in numerator and denominator, we can use mol instead of concentration

we have below equation to be used:

This is Henderson–Hasselbalch equation

pH = pKa + log {[conjugate base]/[acid]}

= 4.8182+ log {0.4464/0.2151}

= 5.14

Answer: 5.14