Question

The number of $S^{2-}$ ions present in 1 L of 0.1 M $H_{2}S$ $[K_{a\left(H_{2}s\right)}={10}^{-21}]$ solution having $\left[H^{\oplus }\right]=0.1$ Mis

• A. $6.023\times {10}^3$
• B. $6.023\times {10}^4$
• C. $6.023\times {10}^5$
• D. $6.023\times {10}^6$

Answer 1

The correct option is A $6.023\times {10}^3$

As we know,

$H_{2}S\to 2H^{+}+S^{-2}$

$K_{a{H}_{2}S}=\left[H^{+}{]}^2\right[S^{-2}\left]/\right[H_{2}S]$

as given,

$\left[S^{-2}\right]=0.1\ast {10}^{-21}/{10}^{-2}={10}^{-20}$

so no of ion of $S^{-2}=6.023\ast {10}^{23}\ast {10}^{-20}=6.02\ast {10}^3$