At 298 K, 0.1 M solution of acetic acid is `1.34 %` ionised . What is the ionisation constant `(K_(a))` for the acid ?

Explanation

Degree of ionisation, `alpha` = 1.34% = 0.0134
Concentration at equilibrium
`CH_3COOH(aq) + H_2O hArr CH_3COO^(-) (aq) + H_3O^(+)`
Since the initial concentration is 0.1 M, the number of moles of acetic acid ionised
0.1 x 0.0134 = 0.00134 mol
Number of moles of acetic acid unionised =0.1 -0.00134 =0.09866 mol
According to the reaction, 1 mol of acetic acid gives 1 mol of `CH_3COO^-` and 1 mol of `H_3O^+` ions. Therefore, 0.00134 mol of `CH_3COOH` will give 0.00134 mol of `CH_3COO^-` and 0.00134 mol of `H_3O^+` ions. Thus, the concentrations at equilibrium are :
`underset"0.09866 mol"(CH_3COOH(aq))+H_2O hArr underset"0.00134 mol"(CH_3COO^(-)(aq))+ underset"0.00134 mol"(H_3O^(+))`
`K_alpha=([CH_3COO^(-)(aq)][H_3O^(+)(aq)])/(CH_3COOH(aq)]`
`=((0.00134)xx(0.00134))/0.09866=1.82xx10^(-5)`