The solubility in water of a sparingly soluble salt $AB_2$ is $1.0 \times 10^{-5} \ mol \ L^{-1}$. Its solubility product constant will be

$A$ solution which is $10^{-3} \ M$ each in $Mn^{2+}, Fe^{2+}, Zn^{2+},$ and $Hg^{2+}$ is treated with $10^{-16} \ M$ sulphide ion. If $K_{sp}$ of $MnS, FeS, ZnS,$ and $HgS$ are $10^{-15}, 10^{-23}, 10^{-20},$ and $10^{-54}$ respectively,which one will precipitate first?

Explain the equilibrium involving the dissolution of a solid in a liquid.

If the solubility product $K_{sp}$ of a sparingly soluble salt $MX_2$ at $25\,^{\circ}C$ is $1.0 \times 10^{-11}$,the solubility of the salt in $\text{mol L}^{-1}$ at this temperature will be:

Solubility product is

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