At $25\,^oC\,,\, K_{sp}$ for $PbBr_2$ is equal to $8 \times 10^{-5}$ . If the salt is $80\%$ dissociated, what is the solubility of $PbBr_2$ in $mol/litre$ ?

The solubility product of $PbI _{2}$ is $8.0 \times 10^{-9} .$ The solubility of lead iodide in $0.1$ molar solution of lead nitrate is $x \times 10^{-6} \,mol / L$. The value of $x$ is ……. .(Rounded off to the nearest integer)

$[$ Given $: \sqrt{2}=1.41]$

Solubility product of a salt $AB $ is $1 \times {10^{ – 8}}$ in a solution in which concentration of $A$ is ${10^{ – 3}}\,M$. The salt will precipitate when the concentration of $B$  becomes more than

In an aqueous solution $SCN^{-1}, B^{-1}, I^{-1}$ and $Cl^-$ are present. Which will get precipitated first, when $AgNO_3$ is mixed with each of them? Given that 
$Ksp$ of $AgCl = 1.2 \times 10^{-10}$,
$Ksp$ of $AgI = 1.7 \times 10^{-16}$
$Ksp$ of $AgSCN = 7.1 \times 10^{-7}$,
$Ksp$ of $AgBr = 3.5 \times10^{-13}$

Solubility of $Pb{I_2}$ is $0.005\,\, M$. Then, the solubility product of $Pb{I_2}$ is