$0.1$ $mol$ of $H_2S(g)$ is kept in a $0.4$ litre vessel at $1000\,K$. For the reaction -
$2{H_2}S(g)\,\rightleftharpoons\,2{H_2}(g)\, + \,{S_2}(g)\,;\,{K_c} = {10^{ - 6}}\% $ dissociation of $H_2S$ is.......$\%$

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

$0.01$ moles of a weak acid $HA \left( K _{ a }=2.0 \times 10^{-6}\right)$ is dissolved in $1.0\, L$ of $0.1\, M\, HCl$ solution. The degree of dissociation of $HA$ is ............. $\times 10^{-5}$

(Round off to the Nearest Integer).

[Neglect volume change on adding $HA$. Assume degree of dissociation $<< 1]$

Degree of dissociation of $0.1\,N\,\,C{H_3}COOH$ is (Dissociation constant $ = 1 \times {10^{ - 5}}$)

Ionic product of water at $310 \,K$ is $2.7 \times 10^{-14}$. What is the $\mathrm{pH}$ of neutral water at this temperature?

The first ionization constant of $H _{2} S$ is $9.1 \times 10^{-8}$. Calculate the concentration of $HS ^{-}$ ion in its $0.1 \,M$ solution. How will this concentration be affected if the solution is $0.1\, M$ in $HCl$ also? If the second dissociation constant of $H _{2} S$ is $1.2 \times 10^{-13}$, calculate the concentration of $S^{2-}$ under both conditions.