Nicotinic acid `(K_(a)=1.4xx10^(-5))` is repersented by the formula HNiC. Calculate its per cent dissociation in a solution, which contains `0.10` mole of nicotinic acid per `2.0` litre of solution.

Explanation

Suppose degree of dissociation of nicotinic acid = `alpha`
`{:(,HNiC,hArr,H^(+),+,NiC^(-)),("Initial amount",0.1 ",mole",,,,),("Amount",0.1-0.1 alpha,,0.1 alpha,,0.1 alpha),("at eqm.",=0.1 (1-alpha),,,,),("Molar conc.",=0.1(1-alpha)//2,,0.1 alpha//2,,0.1 alpha//2),("at eqm.",,,,,):}`
`K_(a)=([H^(+)][NiC^(-)])/([HNiC])`
or `1.4xx10^(-5)=((0.1 alpha)^(2)(0.1alpha//2))/(0.1(1-alpha)//2)`
If `alpha lt lt 1`, then `1.4xx10^(-5)=((0.05 alpha)^(2))/(0.05)=0.05 alpha^(2)`
or `alpha^(2)=2.8xx10^(-4)`
or `alpha=1.67xx10^(-2)`
% dissociation `= 1.67xx106(-2)xx100=1.67%`