The solubility of lithium sodium hexafluoroaluminate $Li_3Na_3(AlF_6)_2$ is $s \ mol \ L^{-1}$. Its solubility product constant $(K_{sp})$ will be:

The solubility product of a salt with the general formula $MX_2$ is $4.0 \times 10^{-12}$. What is the concentration of the $M^{2+}$ ion in a saturated solution of this salt?

The $K_{sp}$ for $BaSO_4$ at $25\ ^\circ C$ 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 (in $M$):

The solubility of an $MX_2$ type electrolyte is $0.5 \times 10^{-4} \ mol/L$. The value of $K_{sp}$ of the electrolyte is:

At $298 \ K$ temperature,the $K_{sp}$ of $CaF_2$ is $1.7 \times 10^{-10}$. If a person drinks $2.5 \ L$ of $CaF_2$ saturated water daily,how many grams of $CaF_2$ are present in the water consumed? (Molecular mass of $CaF_2$ is $78 \ g \ mol^{-1}$)

The solubility product of $AgI$ at $25^{\circ} C$ is $1.0 \times 10^{-16} \ mol^2 \ L^{-2}$. What is the solubility of $AgI$ in $10^{-4} \ N \ KI$ at $25^{\circ} C$?