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The concentration of sulphide ion in $0.1$ $M$ $HCl$ solution saturated with hydrogen sulphide is $1.0 \times 10^{-19} \,M .$ 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?
Given $K_{s p}$ for $FeS =6.3 \times 10^{-18},$ $MnS =2.5 \times 10^{-13}, ZnS =1.6 \times 10^{-24},$ $CdS =8.0 \times 10^{-27}$
Solution
For precipitation to take place, it is required that the calculated ionic product exceeds the $K_{s p}$ value.
Before mixing:
$\left[ S ^{2-}\right]=1.0 \times 10^{-19} \,M \,\,\,\,\left[ M ^{2+}\right]=0.04 \,M$
volume $=10\, mL \quad \quad \quad \,\,$ volume $=5 \,mL$
After mixing:
$\left[ S ^{2-}\right]=?$ volume $=(10+5)=15\, mL$
$\left[ M ^{2+}\right]=?$ volume $=15 \,mL$
$\left[ S ^{2-}\right]=\frac{1.0 \times 10^{-19} \times 10}{15}=6.67 \times 10^{-20}\, M$
$\left[ M ^{2+}\right]=\frac{0.04 \times 5}{15}=1.33 \times 10^{-2} \,M$
Ionic product $=\left[ M ^{2+}\right]\left[ S ^{2-}\right]$
$=\left(1.33 \times 10^{-2}\right)\left(6.67 \times 10^{-20}\right)$
$=8.87 \times 10^{-22}$
This ionic product exceeds the $K_{s p}$ of $Z n s$ and $C d S$ Therefore, precipitation will occur in $CdCl _{2}$ and $ZnCl _{2}$ solutions.
