The solubility product constants of $M(OH)_3$ and $M(OH)_2$ are $10^{-23}$ and $10^{-14}$ respectively. If both ions are present in a solution,which one will precipitate first upon the addition of $NH_4OH$?

If the solubility product of $Ni(OH)_2$ is $4.0 \times 10^{-15}$, the solubility (in $mol \ L^{-1}$) is

The solubility product of a sparingly soluble salt $AX_2$ is $3.2 \times 10^{-11}$. Its solubility (in moles/litres) is

Select the ion exchange reaction,which proceeds in the forward direction in an aqueous medium.

$A$ $1.0 \ L$ of aqueous solution contains $1 \times 10^{-8} \ M \ NaBr$,$1 \times 10^{-8} \ M \ NaCl$ and $1 \times 10^{-8} \ M \ NaI$. To this solution,$1 \times 10^{-10} \ M$ aqueous $AgNO_3$ solution is added dropwise. The order of precipitation of $AgX$ $(X = Cl, Br, I)$ is:
$(K_{sp}(AgCl) = 1.8 \times 10^{-10}; K_{sp}(AgBr) = 5 \times 10^{-13}; K_{sp}(AgI) = 8.3 \times 10^{-17})$

The solubility of $AgCl$ in $0.1 \text{ M } NaCl$ is $S \text{ mol/L}$. If the solubility product of $AgCl$ is $1.8 \times 10^{-10}$, then $S$ is approximately: