The solubility product of a sparingly soluble salt $BA$ is $6.4 \times 10^{-13}$. Calculate its solubility in $\text{g dm}^{-3}$. The molar mass of the salt is $190 \text{ g mol}^{-1}$.

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 $M$):

Concentration of $H_2SO_4$ and $Na_2SO_4$ in a solution is $1 \ M$ and $1.8 \times 10^{-2} \ M$,respectively. Molar solubility of $PbSO_4$ in the same solution is $X \times 10^{-Y} \ M$ (expressed in scientific notation). The value of $Y$ is. . . .
[Given: Solubility product of $PbSO_4$ $(K_{sp})=1.6 \times 10^{-8}$. For $H_2SO_4$,$K_{a1}$ is very large and $K_{a2}=1.2 \times 10^{-2}$]

Solubility product $(K_{sp})$ of the salts $MX$,$MX_2$,and $M_3X$ at temperature $T$ are $4 \times 10^{-8}$,$3.2 \times 10^{-14}$,and $2.7 \times 10^{-15}$ respectively. The solubilities of these salts at temperature $T$ are in the order:

In which of the following pairs of insoluble chlorides,one dissolves in concentrated $HCl$ and the other is not soluble?

$K_{sp}$ for $CaSO_4$ is $9 \times 10^{-6}$. The minimum volume of water needed to dissolve $1 \ g$ of $CaSO_4$ at $298 \ K$ temperature is ...... (in $L$)