If the solubility product of lead iodide $(PbI_2)$ is $3.2 \times 10^{-8}$,then its solubility in $moles/litre$ will be

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 solubility product of $Mg(OH)_2$ is $1 \times 10^{-12}$. At what $pH$ will precipitation occur in a $0.01 \ M$ $MgCl_2$ solution?

Equal volumes of $0.002 \, M$ solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate?
(For cupric iodate $K_{sp} = 7.4 \times 10^{-8}$)

The solubility of sparingly soluble salts $MX$, $MX_2$ and $MX_3$ is $1 \times 10^{-3} \text{ mol/L}$. Hence, their respective solubility products are

The solubility of $AgBr_{(s)}$,having solubility product $5 \times 10^{-10}$ in $0.2 \ M$ $NaBr$ solution,equals