Question

A 1.0 L solution contains 0.25 M HF and 0.32 M NaF (Ka for HF is 7.2x10^-4.

a) What is the pH of the above solution

b) If one adds 0.3 L of 0.020 M KOH to the solution, what will be the pH of the solution?

Answer 1

The above solution is a buffer

pH = pKa + log [salt / acid]

pKa of HF = 3.15

pH = 3.15 + log [0.32 / 0.25] = 3.15 + log (1.28) = 3.15 + 0.107 = 3.257

b) If we add 0.3 L of 0.02 M KOH

so we are adding moles of KOH = molarity X L = 0.02 X 0.3 = 0.006 moles

The KOH will react with HF to give F- ( salt) and it will consume acid by the same amount

so moles of HF consumed = 0.006

moles of HF present initialy = 0.25 x 1 = 0.25

moles left = 0.25 - 0.006 = 0.244 moles

so moles of F- produced= 0.006

moles of HF present initialy = 0.32 x 1 = 0.32

moles left = 0.32 + 0.006 = 0.326moles

pH = pka = log [salt / acid] = 3.15 + log [0.326 / 0.244] = 3.15 + 0.125 = 3.275