Among different types of sparingly soluble salts having nearly the same solubility product constants $(K_{sp})$,the least soluble salt is that which:

The required amount of $KBr$ (molar mass $= 119 \ g/mol$) in grams to start the precipitation of $AgBr$ in $500 \ mL$ solution of $0.05 \ M \ AgNO_3$ will be :- ($K_{SP}$ of $AgBr = 5 \times 10^{-13}$)

Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$BaC_2O_4 \downarrow + 2AcOH \longrightarrow Ba(AcO)_2 + H_2C_2O_4$

The solubility product of silver bromide is $5.0 \times 10^{-13}$. The quantity of potassium bromide (molar mass taken as $120 \ g \ mol^{-1}$) to be added to $1 \ L$ of $0.05 \ M$ solution of silver nitrate to start the precipitation of $AgBr$ is

The solubility products of $NiS$,$ZnS$,$CdS$,and $HgS$ are $4.7 \times 10^{-5}$,$1.6 \times 10^{-24}$,$8 \times 10^{-27}$,and $4 \times 10^{-53}$ respectively. An aqueous solution contains $Ni^{2+}$,$Zn^{2+}$,$Cd^{2+}$,and $Hg^{2+}$ of equal concentration. $H_2S$ gas was passed into this solution very slowly. The first and the last ions that precipitate as sulphides are respectively.

At $T(K)$ in a saturated solution of $MgCO_3$ and $Ag_2CO_3$, if the concentration of $Mg^{2+}$ ion is $3.2 \times 10^{-5} \ M$, then the concentration of $Ag^{+}$ ion in the solution will be [Given: $K_{sp}(MgCO_3) = 1.6 \times 10^{-6}$ and $K_{sp}(Ag_2CO_3) = 8.0 \times 10^{-12}$ at $T(K)$]