Question

$H_{2}S$ is bubbled into a 0.2 M NaCN solution which is, 0.02 M each in $Ag(CN{)}_2^{⊝}$ and $\left(Cd\right(CN{)}_4{)}^{2-}$. If $K_{sp}$ of $A{g}_{2}S$ and $CdS$ are ${10}^{-50}$ and $7.1\times {10}^{-28}$ and K instability for $\left[Ag\right(CN{)}_2^{⊝}]$ and $\left[Cd\right(CN{)}_4^{2-}]$ are $1.0\times {10}^{-20}$ and $7.8\times {10}^{-18}$, which sulphide will precipitate first?

• A. $CdS$
• B. $A{g}_{2}S$
• C. $N{a}_{2}S$
• D. Either $A{g}_{2}S$ or $N{a}_{2}S$ depending on acidity of solution

Answer 1

The correct option is A $CdS$

$H_{2}S\rightleftharpoons 2H^{\oplus }+S^{2-}$
$NaCN⟶N{a}^{\oplus }+C{N}^{⊝}$
0.2 0 0
0 0.2 0.2

Also $Ag(CN{)}_2^{⊝}\rightleftharpoons A{g}^{\oplus }+2C{N}^{⊝}$

$K=\frac{\left[A{g}^{\oplus }\right]\left[\right[C{N}^{⊝}]}{\left[Ag\right(CN{)}_2]}=\frac{\left[A{g}^{\oplus }\right]\left[\right[(0.2{)}^2}{0.02}=1.0\times {10}^{-20}$

$\left[A{g}^{\oplus }\right]=5\times {10}^{-21}M$

For $Cd(CN{)}_4^{2-}\rightleftharpoons C{d}^{2+}+4C{N}^{⊝}$

$K=\frac{\left[C{d}^{2+}\right]\left[C{N}^{⊝}\right]}{Cd(CN{)}_4^{2-}}$

$7.8\times {10}^{-18}=\frac{\left[C{d}^{2+}\right](0.2{)}^4}{0.02}$

$\left[C{d}^{2+}\right]=9.75\times {10}^{-17}$

Now, $[A{g}^{\oplus }{]}^2\times [S^{2-}]=1.0\times {10}^{-50}$

$\left[S^{2-}\right]$ nedded to precipitate $A{g}_{2}S$

$=\frac{1.0\times {10}^{-50}}{(5\times {10}^{-21}{)}^2}=4\times {10}^{-10}M$

$\left[S^{2-}\right]$ needed to precipitate CdS

$=\frac{7.1\times {10}^{-28}}{(9.75\times {10}^{-17})}=7.28\times {10}^{-12}M$

Therefore, CdS will precipitate first.