If the solubility of lithium sodium hexafluoroaluminate $Li_3Na_3(AlF_6)_2$ is $S \ mol/L$,then its solubility product $K_{sp}$ is equal to?

The solubility product of $NiS$ is $4.9 \times 10^{-5}$ at $298 \ K$. Calculate its solubility in $mol \ dm^{-3}$ at the same temperature?

The solubility product of $Hg_2Cl_2$ in water at $25 \, ^\circ C$ is $3.2 \times 10^{-17} \, mol^3 \, dm^{-9}$. The solubility of $Hg_2Cl_2$ at $25 \, ^\circ C$ is:

Predict whether a precipitate of $PbI_2$ will be formed or not on mixing $20 \ mL$ of $3 \times 10^{-3} \ M$ $Pb(NO_3)_2$ solution with $80 \ mL$ of $2 \times 10^{-3} \ M$ $NaI$ solution. $K_{sp}$ for lead iodide $(PbI_2)$ is $6.0 \times 10^{-9}$.

$K_{sp}$ for $BaSO_4$ at $25 \, ^oC$ is $1.1 \times 10^{-10}$. To make the new solubility equal to $1.1 \times 10^{-8} \, M$,it is necessary to use a solution of $Na_2SO_4$ of the following concentration $....... \ M$

Solubility of a salt $M_2X_3$ is $y \ mol \ dm^{-3}$. The solubility product of the salt will be