Solubility product is

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 will precipitation take place? Given $K_{sp}$ 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 minimum volume of water required to dissolve $0.1 \ g$ of lead $(II)$ chloride to get a saturated solution ($K_{sp}$ of $PbCl_2 = 3.2 \times 10^{-8}$; atomic mass of $Pb = 207 \ u$,$Cl = 35.5 \ u$) is......$L$

What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes,there is no precipitation of iron sulphide? (For iron sulphide,$K_{sp} = 6.3 \times 10^{-18}$).

The solubility of $BaSO_4$ in water is $2.33 \times 10^{-3} \ g/L$. Its solubility product will be (molecular weight of $BaSO_4 = 233$).

The solubility product of a salt $AB$ is $1 \times 10^{-8}$ in a solution in which the concentration of $A$ is $10^{-3} \ M$. The salt will precipitate when the concentration of $B$ becomes more than: