The dissociation constant of a substituted benzoic acid at $25^{\circ} \mathrm{C}$ is $1.0 \times 10^{-4}$. The $\mathrm{pH}$ of a $0.01 \ \mathrm{M}$ solution of its sodium salt is

Derive ${K_a} \times {K_b} = {K_w}$ equation.

The molar conductivity of a solution of a weak acid $HX (0.01\ M )$ is $10$ times smaller than the molar conductivity of a solution of a weak acid $HY (0.10 \ M )$. If $\lambda_{ X }^0 \approx \lambda_{ Y ^{-}}^0$, the difference in their $pK _{ a }$ values, $pK _{ a }( HX )- pK _{ a }( HY )$, is (consider degree of ionization of both acids to be $\ll 1$ )

$K _{ a_1,}, K _{ a_2 }$ and $K _{ a_3}$ are the respective ionization constants for the following reactions $(a), (b),$ and $(c)$.

$(a)$ $H _{2} C _{2} O _{4} \rightleftharpoons H ^{+}+ HC _{2} O _{4}^{-}$

$(b)$ $HC _{2} O _{4}^{-} \rightleftharpoons H ^{+}+ HC _{2} O _{4}^{2-}$

$(c)$ $H _{2} C _{2} O _{4} \rightleftharpoons 2 H ^{+}+ C _{2} O _{4}^{2-}$

The relationship between $K_{a_{1}}, K_{ a _{2}}$ and $K_{ a _{3}}$ is given as

What is the $pH$ of solution of $7$ $gm$ $N{H_4}OH$ per $500$ $mL$ ? ( ${K_b}$ of $N{H_4}OH 1.8 \times {10^{ - 5}}$, Molecular moles of $N{H_4}OH$ is $35\,g\,mo{l^{ - 1}}$ )

The dissociation constant of an acid $HA$  is $1 \times {10^{ - 5}}$. The $pH$ of $0.1$ molar solution of the acid will be