Question

When a 15.2 mL sample of a 0.425 M aqueous hydrocyanic acid solution is titrated with a 0.398 M aqueous barium hydroxide solution, what is the pH after 12.2 mL of barium hydroxide have been added?

Answer 1

The reaction is as follows

2HCN + Ba(OH)₂ Ba(CN)₂ + H₂O

So 1 equivalent of hydrocyanic acid requires 0.5 equivalent of  barium hydroxide

15.2 mL of 0.425 M aqueous hydrocyanic acid has (15.2 x 0.425)/1000 = 6.46 x 10^-3 moles of HCN

12.2 mL of 0.398 M aqueous barium hydroxide has (12.2 x 0.398)/1000 = 4.85 x 10^-3 moles of Ba(OH)₂

as can be seen Ba(OH)₂ in this mixture is more than half equivalent with respect to HCN

Ba(OH)₂ is a strong base and so the pH of the solution will be basic

excess of Ba(OH)₂ at the end of the titration is

4.85 x 10^-3 moles - ( 6.46 x 10^-3 moles/2) = 0.00162 moles in 15.2 mL + 12.2 mL = 27.4 mL

That is (0.00162 moles x 1000)/27.4 mL = 0.059 M

One mole of Ba(OH)₂ forms 1 mole of Ba²⁺(aq) and 2 moles of OH^-(aq)

so OH^- in 0.059 M Ba(OH)₂ is 0.059 x 2 = 0.118 M

pOH = -log[OH^-]

= - log (0.118)

pOH = 0.928

pH = 14 - pOH

pH = 14- 0.928

pH = 13.07