Question

What is the maximum $pH$ required to prevent precipitation of $ZnS$ in a solution that is $0.01MZnC{l}_2$ and saturated with $0.1M{H}_{2}S$?

[Given: $K_{sp}\left(ZnS\right)={10}^{-21}$, for $H_{2}SK{a}_1\times K{a}_2={10}^{-20}]$.

Answer 1

$H_{2}S\rightleftharpoons H^{+}+H{S}^{-}{K}_1=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}$

$H{S}^{-}=H^{+}+S^{2-}{K}_2=\frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H{S}^{-}\right]}$

$K_1\times K_2=\frac{\left[H^{+}{]}^2\right[S^{2-}]}{\left[H_{2}S\right]}={10}^{-20}$

$\left[H_{2}S\right]=0.1M$ (given)

${10}^{-20}=\frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[0.1\right]}$

${10}^{-20}\times 0.1=\left[H^{+}{]}^2\right[S^{2-}]$

$\frac{{10}^{-21}}{\left[S^{2-}\right]}=[H^{+}{]}^2..........(1)$

From the equation we notice for heigh $PH$ values is law come of $H^{+}$, the come od free surface ions is heigh

A straight relationship between $pH$ and common of surface

$ZnC{l}_2\to A{n}^{2+}+2Cl$

$0.01m0.01m0.02m$

$KZns\to Z{n}^{2+}+S^{2-}$

$K_{SP}=\left[Zn\right]\left[S^2\right]$

${10}^{-21}=\left[0.01\right]\left[S^{2-}\right]\Rightarrow \left[S^{2-}\right]={10}^{-19}$

Substuting in $\left(1\right)$

$\frac{{10}^{-21}}{{10}^{-19}}=\left[H^{+}\right]\Rightarrow \left[H^{+}\right]+10\Rightarrow pH=-\log \left(H^{+}\right)\Rightarrow pH=1$