Question

A 44.2 mL sample of a 0.344 M aqueous hydrocyanic acid solution is titrated with a 0.216 M aqueous potassium hydroxide solution. What is the pH after 29.5 mL of base have been added?

Answer 1

pKa of HCN= 9.31,

the reaction between HCN and NaOH is HCN+ NaOH--->NaCN+ H2O

1 mole of HCN reacts with 1mole of NaOH

mole of HCN in the solution = Molarity* Volume (L)= 0.344*44.2/1000=0.0152

moles of NaOH= 0.216*29.5/1000=0.0064

moles of HCN is excess since all the moles of NaOH will be consumed.

excess moles of HCN= 0.0152-0.0064=0.0088

Volume after mixing = 29.5+44.2 = 73.7ml = 73.7/1000 L=0.0737 L

concentration of HCN= 0.0088/0.0737= 0.12

HCN ionizes as HCN+ H2O------->CN- + H3O+

Ka= [CN-] [H3O+]/[HCN]

let x=drop inconcentration of HCN to reach equilibrium

so at equilibrium

[HCN] =0.12-x , [CN-] =[H3O+] =x

Ka= x²/(0.12-x)= 4.9*10^-10

comparing at the magniture of RHS, 0.12-x can be approximated to 0.12

x²/0.12= 4.9*10^-10

x= 7.6*10^-6

pH= 5.11