The solubility of $AgBr$ with solubility product $5.0 \times 10^{-13}$ at $298 \ K$ in $0.1 \ M$ $NaBr$ solution would 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 values of $\Delta G_f^o$ (in $kJ \ mol^{-1}$) for $Mg(OH)_2$ at $25 \ ^oC$ are given below. Calculate the solubility product constant $(K_{sp})$.
$Mg^{2+}_{(aq)} = -456.0; OH^{-}_{(aq)} = -157.3; Mg(OH)_{2(s)} = -833.9$

Increasing order of solubility of $AgCl$ in $(i) H_2O$, $(ii) 1 \ M \ NaCl \ (aq.)$, $(iii) 1 \ M \ CaCl_2 \ (aq.)$ and $(iv) 1 \ M \ NaNO_3 \ (aq.)$ solution.

The expression for solubility product of mercurous iodide $(Hg_2I_2)$ is

The molar solubility (in $mol \cdot L^{-1}$) of a sparingly soluble salt $MX_4$ is $S$. The corresponding solubility product is $K_{sp}$. $S$ in terms of $K_{sp}$ is given by the relation: