Calculate the $pH$ of a $0.02 \ M$ ammonia solution that is $5\%$ ionized.

At $27^{\circ} C$,the degree of dissociation of a weak acid $(HA)$ in its $0.5 \ M$ aqueous solution is $1 \%$. Its $K_{a}$ value is approximately:

Consider the dissociation equilibrium of the following weak acid $HA \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$. If the $pK_{a}$ of the acid is $4$, then the $pH$ of $10 \ mM$ $HA$ solution is . . . . . . . (Nearest integer)
[Given : The degree of dissociation can be neglected with respect to unity]

The $pH$ of a $0.1 \ M$ solution of $NH_{4}OH$ (having $K_{b} = 1.0 \times 10^{-5}$) is equal to

The $pH$ of an aqueous solution of $H_2CO_3$ is $3.3$. If ${K_{a_1}} = {10^{-3}}$ and ${K_{a_2}} = {10^{-13}}$,then the concentration of $[HCO_3^-]$ is:

$0.01 \ M \ HA_{(aq)}$ is $2\%$ ionized; $[OH^{-}]$ of solution is