Question

A 1.00 L buffer solution is composed of 0.250 M HF and 0.250 M NaF. Calculate the 13)______ pH of the solution after the addition of 0.150 moles of solid NaOH. Assume no volume change upon the addition of base. The Ka for HF is 3.5 × 10-4.

Answer 1

mol of NaOH added = 0.15 mol

HF will react with OH- to form F-

Before Reaction:
mol of F- = 0.25 M *1.0 L
mol of F- = 0.25 mol

mol of HF = 0.25 M *1.0 L
mol of HF = 0.25 mol

after reaction,
mol of F- = mol present initially + mol added
mol of F- = (0.25 + 0.15) mol
mol of F- = 0.4 mol

mol of HF = mol present initially - mol added
mol of HF = (0.25 - 0.15) mol
mol of HF = 0.1 mol


Ka = 3.5*10^-4

pKa = - log (Ka)
= - log(3.5*10^-4)
= 3.456

since volume is both in numerator and denominator, we can use mol instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.456+ log {0.4/0.1}
= 4.06

Answer: 4.06