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

$A$ $1.0 \ L$ of aqueous solution contains $1 \times 10^{-8} \ M \ NaBr$,$1 \times 10^{-8} \ M \ NaCl$ and $1 \times 10^{-8} \ M \ NaI$. To this solution,$1 \times 10^{-10} \ M$ aqueous $AgNO_3$ solution is added dropwise. The order of precipitation of $AgX$ $(X = Cl, Br, I)$ is:
$(K_{sp}(AgCl) = 1.8 \times 10^{-10}; K_{sp}(AgBr) = 5 \times 10^{-13}; K_{sp}(AgI) = 8.3 \times 10^{-17})$

The solubility in water of a sparingly soluble salt $AB_2$ is $1.0 \times 10^{-5} \ mol \ L^{-1}$. Its solubility product constant will be

The solubility product of $Ag_{2}CrO_{4}$ is $32 \times 10^{-12}$. What is the concentration of $CrO_{4}^{2-}$ ions in that solution?

Precipitation occurs if the ionic product is ........

Calculate the equilibrium concentration of $Pb^{2+}$ ions in a solution of $PbS$ containing $1 \times 10^{-11} \ mol \ dm^{-3}$ of sulphide ions. (Given $K_{sp}$ for $PbS = 8.0 \times 10^{-28}$)