Question

Two 25.0 ml samples, on 0.100 mol/L HCl and the other 0.100mol/L HF, were titrated with 0.200 mol/L KOH.

a) What is the volume of added base at the equivalence point for each titration?

b) Predict whether the pH at the equivalence point for each titration will be acidic, basic or neutral? Explain your selection.

c) Predict which titration curve will have the lower initial pH and explain your selection.

Answer 1

HCl is more acidic than HF.Therefore HCl will dissociate completely into its respective ions and thus have greater pH value

Since the acids HF & HCl are strong, and also the base KOH is strong, therefore complete neutralisation will occur and hence the solution will be neutral.

Now,

HCl + KOH -------> KCl + H₂O

as per the above balanced reaction, HCl & KOH reacts in the molar ratio of 1:1

Thus, moles of KOH required = moles of HCl = 0.1*0.025 = 0.0025

volume of KOH required = moles/molarity = 0.0025/0.2 = 0.0125 L = 12.5 ml

HCl + KOH -------> KCl + H₂O

as per the above balanced reaction, HF & KOH reacts in the molar ratio of 1:1

Thus, moles of KOH required = moles of HF = 0.1*0.025 = 0.0025

volume of KOH required = moles/molarity = 0.0025/0.2 = 0.0125 L = 12.5 ml