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 $FeS =6.3 \times 10^{-18},$ $MnS =2.5 \times 10^{-13}, ZnS =1.6 \times 10^{-24},$ $CdS =8.0 \times 10^{-27}$

The solubility of $PbC{l_2}$ at ${25\,^o}C$ is $6.3 \times {10^{ – 3}}$ $mole/litre$. Its solubility product at that temperature is

Solubility product is

The solubility product constant ${K_{sp}} $ of $Mg{(OH)_2}$ is $9.0 \times {10^{ – 12}}.$ If a solution is $0.010\,\,M$ with respect to $M{g^{2 + }}$ ion, what is the maximum hydroxide ion concentration which could be present without causing the precipitation of $Mg\,{(OH)_2}$

The concentration of $500$ $ML$ $NaOH$ solution is $0.02$ $M$. How many grams of $FeS{O_4}$ added in this solution for precipitation of $Fe{\left( {OH} \right)_2}$ ? The ${K_{sp}}$ of $Fe{\left( {OH} \right)_2}$ is $1.5 \times {10^{ – 15}}$. Molecular mass of $Fe{\left( {OH} \right)_2}$ is $ 152$ $g$ $mol^{-1}$ )