Question

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

• 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 $6.023\times {10}^3$

Given reaction is,
$H_{2}S\leftrightharpoons 2H^{+}+S^{2-}$
$\therefore K_a=\frac{\left[H^{+}{]}^2\right[S^{2-}]}{\left[H_{2}S\right]}$
$\Rightarrow {10}^{-21}=\frac{(0.1{)}^2\times [S^{2-}]}{0.1}$
$\therefore \left[S^{2-}\right]={10}^{-20}M$
So, total number of ions = ${10}^{-20}\times 6.023\times {10}^{23}=6.023\times {10}^3$ ions