Degree of dissociation of $0.1\,N\,\,C{H_3}COOH$ is (Dissociation constant $ = 1 \times {10^{ - 5}}$)
${10^{ - 5}}$
${10^{ - 4}}$
${10^{ - 3}}$
${10^{ - 2}}$
The $pH$ value of decinormal solution of $N{H_4}OH$ which is $20\%$ ionised, is
Calculate $pH$ of $0.02$ $mL$ $ClC{H_2}COOH$. Its ${K_a} = 1.36 \times {10^{ - 3}}$ calculate its $pK_{b}$,
$0.01\, M \,HA(aq.)$ is $2\%$ ionized, $[OH^-]$ of solution is :-
The dissociation constants of two acids $HA_1$ and $HA_2$ are $3.0 \times 10^{-4}$ and $1.8 \times 10^{-5}$ respectively. The relative strengths of the acids will be
What is the percent ionization $(\alpha)$ of a $0.01\, M\, HA$ solution ? .......$\%$ $(K_a = 10^{-6})$