Question

What $\left[H_3{O}^{\oplus }\right]$ must be maintained in a saturated $H_{2}S$ solution to precipitate $P{b}^{2+}$ but not $Z{n}^{2+}$ from a solution in which each ion is 0.01 M? $K_{sp}\left(H_{2}S\right)=1.1\times {10}^{-22},K_{sp}\left(ZnS\right)=1.0\times {10}^{-21}$

• A. $3.3\times {10}^{-2}M$
• B. $3.3\times {10}^{-5}M$
• C. $3.3\times {10}^{-1}M$
• D. $3.3M$

Answer 1

The correct option is A $3.3\times {10}^{-2}M$

For ZnS not to be precipitated from a solution $Z{n}^{2+}$ and $P{b}^{2+}$
$\left[Z{n}^{2+}\right]\left[S^{2-}\right]<K_{sp}$ of ZnS
$\left({10}^{-2}\right)\left[S^{2-}\right]<1.0\times {10}^{-21}$
or the maximum $\left[S^{2-}\right]={10}^{-19}$ at which ZnS will begin to precipitate or upto this concentration, no precipitaion will occur.
$H_{2}S\rightleftharpoons 2H^{\oplus }+S^{2-}$
$\therefore \left[H^{\oplus }{]}^2\right[S^{2-}]=1.1\times {10}^{-22}$
$\therefore \left[H^{+}{]}^2\right[{10}^{-19}]=1.1\times {10}^{-22}$
$\therefore [H^{\oplus }{]}^2=11\times {10}^{-4}\therefore [H^{\oplus }]=3.3\times {10}^{-2}M$
This, is $\therefore \left[H^{\oplus }\right]=3.3\times {10}^{-2}$ or slightly higher, the precipitation of ZnS will not take place and only PbS will precipitate