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
$2.0 \times {10^{ - 6}}\,mol\,{L^{ - 1}}$
$1.0 \times {10^{ - 5}}\,mol\;{L^{ - 1}}$
$1.0 \times {10^{ - 6}}\,mol\,{L^{ - 1}}$
$1.0 \times {10^{ - 7}}\,mol\;{L^{ - 1}}$
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
If the dissociation constant of an acid $HA$ is $1 \times {10^{ - 5}},$ the $pH$ of a $ 0.1$ molar solution of the acid will be approximately
$0.01\, M \,HA(aq.)$ is $2\%$ ionized, $[OH^-]$ of solution is :-
In $20\,\, ml \,\,0.4 \,M-HA$ solution, $80\,\, ml$ water is added. Assuming volume to be additive, the $pH$ of final solution is
$(K_a \,\,of\,\, HA = 4 \times 10^{-7} ,\, log\,2 = 0.3)$
The $pH$ of $0.004 \,M$ hydrazine solution is $9.7 .$ Calculate its ionization constant $K_{ b }$ and $pK _{ b }$