At $298 \ K$,the solubility product of $PbCl_2$ is $1.0 \times 10^{-6}$. What will be the solubility of $PbCl_2$ in $mol \ L^{-1}$?

Which of the following happens when $NH_{4}OH$ is added gradually to the solution containing $1 \ M \ A^{2+}$ and $1 \ M \ B^{3+}$ ions?
Given: $K_{sp}[A(OH)_{2}] = 9 \times 10^{-10}$ and $K_{sp}[B(OH)_{3}] = 27 \times 10^{-18}$ at $298 \ K$.

If the solubility of $PbBr_2$ is $S \, mol/L$,and it undergoes $100\%$ ionization,then the solubility product constant $(K_{sp})$ is equal to:

In an aqueous solution,$SCN^-$,$Br^-$,$I^-$,and $Cl^-$ are present. Which will get precipitated first when $AgNO_3$ is added to the solution? Given that:
$K_{sp}(AgCl) = 1.2 \times 10^{-10}$,
$K_{sp}(AgI) = 1.7 \times 10^{-16}$,
$K_{sp}(AgSCN) = 7.1 \times 10^{-7}$,
$K_{sp}(AgBr) = 3.5 \times 10^{-13}$

The expression for the solubility product of $Al_2(SO_4)_3$ is

If the solubility product constant $(K_{sp})$ of $Ni(OH)_2$ is $1.9 \times 10^{-15}$,the molar solubility of $Ni(OH)_2$ in $1.0 \ M \ NaOH$ is: