Calculate the degree of dissociation and concentration of `H_(3)O^(+)` ions in 0.01 M solution of formic acid `(K_(c) =2.1 xx 10^(-4) " at " 298 K)`

Explanation

Formic acid is weak electrolyte and ionizes in water to give `H_(3)O^(+)` ions according to the equation :
`HCO OH+H_(2)O hArr H_(3)O^(+)+HC O O^(-)`
Let `alpha` be the degree of ionization. Then the concentration of the various species present at equilibrium would be as under :
`{:(,HCO OH + H_(2)O hArrH_(3)O^(+) +HCO O^(-)),("Initial conc. ",0.01 " 0 0"),("Conc. at eqm",0.01(1-alpha)~=0.01 " "0.01 alpha " "0.01alpha):}` [`:' alpha` is very small and can be neglected in comparison to 1]
Thus, `K_(a)=(0.01 alpha xx 0.01 alpha)/(0.01)=0.01 xx alpha^(2) :. 0.01 alpha^(2)=2.1xx10^(-4) [K_(a)=2.1xx10^(-4), "given"]`
or `alpha^(2)=(2.1xx10^(-4))/(0.01)=2.1xx10^(-2)`
`:.` Degree of ionisation, `alpha = sqrt(2.1xx10^(-2))=0.14`
Concentration of `H_(3)O^(+)` ions `=C alpha = 0.14 xx 0.01 = 1.4 xx 10^(-3) "mol" L^(-1)`