The solubility of $CaF_2$ is $2 \times 10^{-4} \, mol/L$. Its solubility product $(K_{sp})$ is:

Calculate the minimum concentration of sulphate ion required to precipitate $BaSO_4$ in a solution containing $10^{-4} \ mol/L$ of $Ba^{2+}$. $(K_{sp} \ BaSO_4 = 4 \times 10^{-10})$

If the solubility product of $BaSO_4$ is $1.5 \times 10^{-9}$ in water,its solubility in moles per litre is

Assertion : Addition of silver ions to a mixture of aqueous sodium chloride and sodium bromide solution will first precipitate $AgBr$ rather than $AgCl$.
Reason : $K_{sp}$ of $AgCl < K_{sp}$ of $AgBr$.

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 $K_{a2}=1.2 \times 10^{-2}$]

The $K_{sp}$ of $AgBr$ at $25 \, ^\circ C$ is $4.9 \times 10^{-13}$. If the molar mass of $AgBr$ is $188 \, g/mol$,how much $AgBr$ is required to saturate $20 \, L$ of water?