At $25^{\circ} C$,the solubility product of $MCl$ is $1 \times 10^{-10}$. What is its molar solubility in $0.1 \ M \ NaCl$ solution at the same temperature?

If $K_{sp}$ is the solubility product of $Al(OH)_3$,its solubility is expressed by the formula:

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

The solubility product of $BaCl_2$ is $4 \times 10^{-9}$. Its solubility in $mol/L$ is .......

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 $K_{sp}$ values of $Ag_2CrO_4, AgCl, AgBr,$ and $AgI$ are $1.1 \times 10^{-12}, 1.8 \times 10^{-10}, 5.0 \times 10^{-13},$ and $8.3 \times 10^{-17}$ respectively. If $AgNO_3$ solution is added to a solution containing equal moles of $NaCl, NaBr, NaI,$ and $Na_2CrO_4$,which will precipitate last?