Find out the solubility of $Ni(OH)_2$ in $0.1 \ M \ NaOH$.
Given that the solubility product constant $(K_{sp})$ of $Ni(OH)_2$ is $2 \times 10^{-15}$.

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:

$Mg(OH)_2$ is precipitated when $NaOH$ is added to a solution of $Mg^{2+}$. If the final concentration of $Mg^{2+}$ is $10^{-10} \ M$,the concentration of $OH^{-} \ (M)$ in the solution is (Given: Solubility product for $Mg(OH)_2 = 5.6 \times 10^{-12}$)

At $T(K)$,$K_{sp}$ of two ionic salts $MX_2$ and $MX$ is $5 \times 10^{-13}$ and $1.6 \times 10^{-11}$ respectively. The ratio of molar solubility of $MX_2$ and $MX$ is

Calculate the molar solubility of $Ni(OH)_{2}$ in $0.10 \, M$ $NaOH$. The solubility product constant $(K_{sp})$ of $Ni(OH)_{2}$ is $2.0 \times 10^{-15}$.

The solubility product of $BaCl_2$ is $4 \times 10^{-9}$. Its solubility in moles/litre would be: