Question

20.0 ml of 0.10M HCN is titrated with 0.10 M KOH. determine the ph at the equivalence point.

Answer 1

This is the case of titration of strong base with weak acid

At equlvalence point,

concentraion of HCN=concentration of KOH

therefore,M₁V₁=M₂V₂

20*0.1=V*0.1

thus, V=20 ml

total volume of solution =20+20=40 ml

here the salt CN^- will be formed whose moles =20.*0.1=2 millimoles

but total volume=40ml

so [CN^-] salt =2/40=0.05 M

The CN^- is a weak base

CN^-+H₂O-->HCN+OH^-

The k_b for CN^- can be calculated from Ka value as KaKb=Kw

Ka for HCN=6.2*10^-10

thus K_b=10^-14/6.2*10^-10=1.612*10^-5

now K_b=x.x/(0.05-x)

assume 0.05>>x so 0.05-x=0.05

thus x=[OH^-]=9*10^-4

thus pOH=-log([OH^-])=3.04

pH=14-3.04=10.95