At  $25\,^o C$, the dissociation constant of a base $BOH$ is $1.0 \times {10^{ - 12}}$. The concentration of Hydroxyl ions in $0.01\, M$ aqueous solution of the base would be

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

${K_a}$ of $C{H_3}COOH$ is $1.76 \times {10^{ - 5}}$ at $298$ $K$ temperature. Calculate dissociation constant of its conjugate base.

What is $[{H^ + }]$ of a solution that is $0.01\,M$ in $HCN$ and $0.02\,M$ in $NaCN$ $({K_a}$for $HCN = 6.2 \times {10^{ - 10}})$

The $pH$ of a $0.1\  M$ aqueous solution of a very weak acid $(HA)$ is $3$. What is its degree of dissociation ?......$\%$

${K_b}$ 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.