The $K_{SP}$ of $AgI$ is $1.5 \times 10^{-16}$. On mixing equal volumes of the following solutions,precipitation will occur only with:

In qualitative analysis,the metals of group $I$ can be separated from other ions by precipitating them as chloride salts. $A$ solution initially contains $Ag^{+}$ and $Pb^{2+}$ at a concentration of $0.10 \, M$. Aqueous $HCl$ is added to this solution until the $Cl^{-}$ concentration is $0.10 \, M$. What will the concentrations of $Ag^{+}$ and $Pb^{2+}$ be at equilibrium? ($K_{sp}$ for $AgCl = 1.8 \times 10^{-10}$,$K_{sp}$ for $PbCl_2 = 1.7 \times 10^{-5}$)

If the solubility product of $AgCl$ is $10^{-10}$,what is the solubility of $AgCl$ in $0.001 \ M \ NaCl$?

The solubility product of the sparingly soluble salt $AB_2$ is $2.56 \times 10^{-10}$ at $298 \ K$. Calculate its solubility in $mol \ dm^{-3}$ at the same temperature?

The solubility product of $BaCl_2$ is $4 \times 10^{-9}$. Its solubility in moles/litre would be:

The solubility product of $AgI$ at $25\,\text{°C}$ is $1.0 \times 10^{-16}\,\text{mol}^2\,\text{L}^{-2}$. The solubility of $AgI$ in $10^{-4}\,\text{N}$ solution of $KI$ at $25\,\text{°C}$ is approximately (in $\text{mol L}^{-1}$)