A 12.4 mL solution of 0.100 mol L-1 HF is titrated using 0.150 mol L-1 NaOH. What is the pH of the solution after 5.61 mL of the NaOH solution is added? Express your answer to 2 decimal places.

Answer 1

HF + NaOH ==> NaF + H2O

moles HF initially present = 12.4 ml x 1 L / 1000 ml x 0.100 mol/L = 0.00124 moles

moles NaOH initially present = 5.61 ml x 1 L / 1000 ml x 0.150 mol/L = 0.000842 mols

After reaction:

moles HF = 0.00124 - 0.000842 = 0.000398 mols HF

moles F- formed = 0.000842 mols F-

This forms a buffer of a weak acid (HF) and the conjugate base (F-)

pH = pKa + log [conj.base] / [acid] ... looking up pKa for HF, we find 3.14

pH = 3.14 + log 0.000842 / 0.000398 = 3.14 + 0.33

pH = 3.47