Question

Compared to K for the dissociation,$2H_{2}S\rightleftharpoons 2H^{+}+2H{S}^{-}$, then $K^{{}^{\prime }}$ for the $H^{+}+H{S}^{-}\rightleftharpoons H_{2}S$ would have:

• A. $\frac{1}{\sqrt{K}}$
• B. $\frac{1}{2K}$
• C. $\frac{1}{2\sqrt{K}}$
• D. $\frac{1}{K}$

Answer 1

The correct option is A $\frac{1}{\sqrt{K}}$

Compared to K for the dissociation,$2H_{2}S\rightleftharpoons 2H^{+}+2H{S}^{-}$, then $K^{{}^{\prime }}$ for the $H^{+}+H{S}^{-}\rightleftharpoons H_{2}S$ would have $\frac{1}{\sqrt{K}}$
First reaction is reversed and divided by 2 to obtain second reaction.
$K={\frac{\left[H^{+}{]}^2\right[H{S}^{-}]}{H_{2}S}}^2$
$K^{\prime }=\frac{\left[H_{2}S\right]}{\left[H^{+}\right]\left[H{S}^{-}\right]}=\frac{1}{\sqrt{{\frac{\left[H^{+}{]}^2\right[H{S}^{-}]}{\left[H_{2}S\right]}}^2}}=\frac{1}{\sqrt{K}}$