At $20 \, ^\circ C$,the concentration of $Ag^+$ ions in a saturated solution of $Ag_2CrO_4$ is $1.5 \times 10^{-4} \, M$. The solubility product $(K_{sp})$ of $Ag_2CrO_4$ at $20 \, ^\circ C$ is:

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

If the concentration of $Cr_2O_7^{2-}$ in a saturated solution of $Ag_2Cr_2O_7$ is $6.5 \times 10^{-5} \ M$ at a constant temperature,calculate the value of $K_{sp}$ for $Ag_2Cr_2O_7$.

In an aqueous solution,$SCN^-$,$Br^-$,$I^-$,and $Cl^-$ are present. Which will get precipitated first when $AgNO_3$ is added to the solution? Given that:
$K_{sp}(AgCl) = 1.2 \times 10^{-10}$,
$K_{sp}(AgI) = 1.7 \times 10^{-16}$,
$K_{sp}(AgSCN) = 7.1 \times 10^{-7}$,
$K_{sp}(AgBr) = 3.5 \times 10^{-13}$

At $298 \ K$ temperature,the solubility product $(K_{sp})$ for $AgCl$ is $1.5 \times 10^{-10}$. Calculate its solubility in $g \ L^{-1}$ in pure water. (Molar mass of $AgCl = 143.5 \ g \ mol^{-1}$)

Two salts $A_2X$ and $MX$ have the same value of solubility product of $4.0 \times 10^{-12}$. The ratio of their molar solubilities i.e. $\frac{S(A_2X)}{S(MX)} = \dots$ (Round off to the Nearest Integer).