Q

# The concentration of sulphide ion in 0. 1M HCl solution saturated with hydrogen sulphide is 1.0 × 10^–19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO_4

7.73     The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10-19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: $FeSO_{4},MnCl_{2},ZnCl_{2} and CdCl_{2}.$ in which of these solutions precipitation will take place?

Views

We have,
the concentration of $[S^{2-}]=1\times 10^{-19}$ and the volume of the solution containing sulphur ion = 10 mL.
Volume of metal salts solution added = 5mL
Before mixing,

$[S^{2-}]=1\times 10^{-19}$
$[M^{2+}]=0.04 M$

After mixing,

Volume = 15 mL

So, the concentration of $[S^{2-}]=1\times 10^{-19}\times \frac{10}{15}$
$=6.67\times 10^{-20}M$

concentration of $[M^{2+}]=0.04 M\times \frac{5}{15}$
$=1.33\times 10^{-2}$

So, the ionic product = $[M^{2+}][S^{2-}]$
= $(6.67 \times 10^{-20})(1.33\times 10^{-2})$
= $8.87 \times 10^{-22}$

For the precipitation of the solution, the ionic product should be greater than the corresponding $K_s_p$ values.
$K_s_p$ of $FeS,\ MnS,\ ZnS,\ CdS$ are $6.3\times 10^{-18},\2.5\times 10^{-13},\ 1.6\times 10^{-24}$ and $8\times 10^{-27}$ respectively.

Hence precipitation will take place in $CdCl_2$ and $ZnCl_2$metal salts.

Exams
Articles
Questions