An aqueous solution contains 0477 M hydrocyanic acid Calcula

An aqueous solution contains 0.477 M hydrocyanic acid.

Calculate the pH of the solution after the addition of 1.55E-2 moles of potassium hydroxide to 125 mL of this solution.
(Assume that the volume does not change upon adding potassium hydroxide).

Solution

pk for HCN = 9.21 , moles of HCN = 0.477x0.125= 0.0596,

moles of KOH added = 0.0155,

HCN + KOH <---> KCN + H2O , KCN formed = 0.0155 moles = (0.0155/0.125),

[HCN} = ( 0.059-0.0155)/0.125 = 0.0441/0.125

pH =pKa + log[KCN]/[HCN]

pH = 9.21 + log(0.0155/0.125)/(0.0441/0.125)

pH = 8.756