What is the pH at the equivalence point in the titration of a 17.1 mL sample of a 0.331 M aqueous hydrocyanic acid solution with a 0.421 M aqueous sodium hydroxide solution?

Question

What is the pH at the equivalence point in the titration of a 17.1 mL sample of a 0.331 M aqueous hydrocyanic acid solution with a 0.421 M aqueous sodium hydroxide solution?

Answer 1

The reaction is between a weak acid (HCN) and a strong base (NaOH). Therefore, at equivalence the pH of the solution will be basic (pH>7).

HCN + NaOH ==> NaCN + H2O

moles HCN present =17.1 ml x 1 L/1000 ml x 0.331 mol/L = 5.66x10^-3 mols

moles NaOH present = 5.66x10^-3 mols in order to reach equivalence

Volume of NaOH need for 5.66x10^-3 mol: 0.421 mol/L (x L) = 5.66x10^-3 mol and x = 0.0134 L

Final volume at equivalence = 0.0171 L + 0.0134 L = 0.0305 L

At equivalence, all the HCN and NaOH have reacted to form 5.6x10^-3 mols of NaCN

CN- + H2O ==> HCN + OH-

Kb = [HCN][OH-]/[CN-] = (x)(x)/5.66x10^-3 - x

Look up Kb for CN- (or calculate it from Ka of HCN), solve for x

x will be [OH-]. Calculate pOH and then pH