Question

If the concentration of lead iodide in its saturated solution be $2\times {{10}^{-3}}$ moles per litre, then its solubility product is:
A. \[4\times {{10}^{-6}}\]
B. \[8\times {{10}^{-12}}\]
C. \[6\times {{10}^{-9}}\]
D. \[32\times {{10}^{-9}}\]

Answer 1

Hint: Solubility product depends on the number of ions formed in a reaction or the concentration of the ions at equilibrium. Lead nitrate dissociates into two lead cations and nitrate anion. Thus the solubility product depends on concentration of the cation and anion.

Complete Step by Step Answer:
Solubility product of a species is defined as the product of the concentration of the cation and anion formed from a species at equilibrium. At this stage the coefficients before the ions exist as a power to the concentration of the corresponding cation and anion.

Lead nitrate dissociate at equilibrium as follows:
$Pb{{(N{{O}_{3}})}_{2}}\rightleftharpoons P{{b}^{+2}}+2N{{O}_{3}}^{-}$
Thus lead nitrate dissociates into one lead cation and two nitrate anions at equilibrium.

Thus solubility product is given as follows:
$K_{sp}={{(2S)}^{2}}\times S$; Where $K_{sp}$ is defined as the solubility product and $S$ is defined as the solubility or concentration for the cation and anion. Solubility of nitrate anion has a coefficient of $2$ .
The concentration of lead nitrate is $2\times {{10}^{-3}}$. Putting the value of solubility in the above equation we get:
$K_{sp}=4\times {{(2\times {{10}^{-3}})}^{3}}$
$K_{sp}=32\times {{10}^{-9}}$ moles per litre
Thus the value of solubility product of lead nitrate is $32\times {{10}^{-9}}$moles per litre.
Thus the correct option is D.

Note: The chemical formula of lead nitrate is $Pb{{(N{{O}_{3}})}_{2}}$ . It is an ionic compound. It dissociates into the corresponding ions in solution state. It is a white colored solid and it is insoluble in water.