${K_b}$ of $N{H_4}OH = 1.8 \times {10^{ - 5}}$ calculate $pH$ of $0.15$ $mol$ $N{H_4}OH$ and $0.25$ $mol$ $N{H_4}OH$ containing solution.

The $pH$ of two equimolar weak acids are $3.0$ and $5.0$ respectively. Their relative strength is

If the $pKa$ of lactic acid is $5$,then the $pH$ of $0.005$ $M$ calcium lactate solution at $25^{\circ}\,C$ is $........\times 10^{-1}$ (Nearest integer)

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 $\ll 1$ )

Given

$(i)$ $\begin{gathered}
  HCN\left( {aq} \right) + {H_2}O\left( l \right) \rightleftharpoons {H_3}{O^ + }\left( {aq} \right) + C{N^ - }\left( {aq} \right) \hfill \\
  {K_a} = 6.2 \times {10^{ - 10}} \hfill \\ 
\end{gathered} $

$(ii)$ $\begin{gathered}
  C{N^ - }\left( {aq} \right) + {H_2}O\left( l \right) \rightleftharpoons HCN\left( {aq} \right) + O{H^ - }\left( {aq} \right) \hfill \\
  {K_b} = 1.6 \times {10^{ - 5}} \hfill \\ 
\end{gathered} $

These equilibria show the following order of the relative base strength

Sulphurous acid $\left( H _{2} SO _{3}\right)$ has $Ka _{1}=1.7 \times 10^{-2}$ and $Ka _{2}=6.4 \times 10^{-8} .$ The $pH$ of $0.588 \,M\, H _{2} SO _{3}$ is ..... . (Round off to the Nearest Integer)