$5\%$ ionization is occur in $0.01$ $M$ $C{H_3}COOH$ solution. Calculate its dissociation constant.

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$2.63 \times 10^{-5}$

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When $CO_2$ dissolves in water, the following equilibrium is established

$C{O_2} + 2{H_2}O\, \rightleftharpoons {H_3}{O^ + } + HCO_3^ - $

for which the equilibrium constant is $3.8 \times 10^{-7}$ and $pH = 6.0$. The ratio of  $[HCO_3^- ]$ to $[CO_2]$ would be :-

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$(i)$ $\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 \\
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\end{gathered} $

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  C{N^ - }\left( {aq} \right) + {H_2}O\left( l \right) \rightleftharpoons HCN\left( {aq} \right) + O{H^ - }\left( {aq} \right) \hfill \\
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