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

In an aqueous solution containing $AgCl_{(s)}$,$AgBr_{(s)}$,and $AgI_{(s)}$,find the ratio $[Cl^{-}] : [Br^{-}] : [I^{-}]$ at equilibrium if $K_{sp}(AgCl) = 10^{-10}$,$K_{sp}(AgBr) = 10^{-13}$,and $K_{sp}(AgI) = 10^{-16}$.

Derive the solubility product $(K_{sp})$ expression for the following sparingly soluble salts:
$(i)$ Two ions having $MX$ formula
$(ii)$ Three ions having $MX_{2}$ or $M_{2}X$ types
$(iii)$ Four ions having $AX_{3}$ or $A_{3}X$ type salts
$(iv)$ Five ions $A_{2}X_{3}$ or $A_{3}X_{2}$ type salts.

$A$ precipitate is formed when:

For $AgCl$,$K_{sp} = 1.2 \times 10^{-10}$ and for $AgBr$,$K_{sp} = 3.5 \times 10^{-13}$. What is the relationship between their solubilities $(S)$?

The solubility products of $Al(OH)_3$ and $Zn(OH)_2$ are $8.5 \times 10^{-23}$ and $1.8 \times 10^{-14}$ at room temperature. If the solution contains $Al^{3+}$ and $Zn^{2+}$ ions,the ion first precipitated by adding $NH_4OH$ is