Question

The ${\text{Ksp}}$ of lead iodide ${\text{Pb}}{{\text{I}}_{\text{2}}}$ is $1.4\, \times {10^{ - 8}}$. Calculate the solubility of lead iodide in each of the following:
(a) water
(b) ${\text{0}}{\text{.10}}$M${\text{Pb}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$
(c) ${\text{0}}{\text{.010}}$M${\text{NaI}}$

Answer 1

Hint: ${{\text{K}}_{{\text{sp}}}}$ represents the solubility product constant at equilibrium. When a solid substance dissolves into water it dissociates into ion. The product of solubilities of the ions with the number of each ion in power is known as the solubility product.

Formula used: ${{\text{K}}_{{\text{sp}}}}\, = \,x{S^x}\, \times \,y{S^y}$

Complete step-by-step solution:
The solubility product constant represents the product of concentrations of constituting ions of an ionic compound at equilibrium each raised number of ions as power.
An ionic compound dissociates into the water as follows:
${{\text{A}}_{\text{x}}}{{\text{B}}_{\text{y}}}\, \rightleftarrows \,\,{\text{x}}{{\text{A}}^{{\text{ + Y}}}}\,{\text{ + }}\,{\text{y}}{{\text{B}}^{ - x}}$
Where, ${\text{AB}}$ is an ionic compound.

The general representation for the solubility product of an ionic compound is as follows:
$\Rightarrow {{\text{K}}_{{\text{sp}}}}\, = \,{\left( {x{A^{ + y}}\,} \right)^x}\, \times \,\,{\left( {y{A^{ - x}}\,} \right)^y}$
$\Rightarrow {{\text{K}}_{{\text{sp}}}}\, = \,x{S^x}\, \times \,y{S^y}$
Where,
${{\text{K}}_{{\text{sp}}}}$ is the solubility product constant.
${\text{S}}$ is the solubility of each ion.
$x$ and $y$ represents the number of ions.

The compound lead iodide ${\text{Pb}}{{\text{I}}_{\text{2}}}$ dissociates into the water as follows:
${\text{Pb}}{{\text{I}}_{\text{2}}}\, \rightleftarrows \,{\text{P}}{{\text{b}}^{{\text{2 + }}}}{\text{ + }}\,2\,{{\text{I}}^ - }$
Chromium hydroxide in water produces one lead ion and two iodide ions.

The solubility product for lead iodide is represented as follows:
$\Rightarrow {{\text{K}}_{{\text{sp}}}}\, = \,{\text{P}}{{\text{b}}^{{\text{2 + }}}}{\text{ + }}\,\,2{{\text{I}}^ - }$
$\Rightarrow {{\text{K}}_{{\text{sp}}}}\, = \,\left( {{\text{S}}\,} \right)\, \times \,\,{\left( {2\,{\text{S}}} \right)^2}$
$\Rightarrow {{\text{K}}_{{\text{sp}}}}\, = \,4{{\text{S}}^3}$
Rearrange the equation for solubility as follows:
$\Rightarrow \,{{\text{S}}^3} = \,\,\dfrac{{{{\text{K}}_{{\text{sp}}}}}}{4}$
$\Rightarrow \,{\text{S}} = \,\,\sqrt[3]{{\dfrac{{{{\text{K}}_{{\text{sp}}}}}}{4}}}$
Substitute $1.4\, \times {10^{ - 8}}$ for ${{\text{K}}_{{\text{sp}}}}$.
$\Rightarrow \,{\text{S}} = \,\,\sqrt[3]{{\dfrac{{1.4\, \times {{10}^{ - 8}}}}{4}}}$
$\Rightarrow \,{\text{S}} = \,\,1.5\, \times {10^{ - 9}}$
So, the solubility of ${\text{Pb}}{{\text{I}}_{\text{2}}}$ in water is$1.5\, \times {10^{ - 9}}$.
The compound lead nitrate ${\text{Pb}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$ dissociates as follows:
$\mathop {{\text{Pb}}{{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)}_{\text{2}}}}\limits_{0.10} \, \rightleftarrows \,\mathop {{\text{P}}{{\text{b}}^{{\text{2 + }}}}}\limits_{0.10} {\text{ + }}\,2\,\mathop {{\text{NO}}_3^ - }\limits_{0.10} $
lead nitrate produces lead ion so, the concentration of lead ion is $0.10$M.

The solubility product for lead iodide is represented as follows:
$\Rightarrow {{\text{K}}_{{\text{sp}}}}\, = \,{\text{P}}{{\text{b}}^{{\text{2 + }}}}{\text{ + }}\,\,2{{\text{I}}^ - }$
$\Rightarrow {{\text{K}}_{{\text{sp}}}}\, = \,\left( {{\text{S + 0}}{\text{.10}}\,} \right)\, \times \,\,{\left( {2\,{\text{S}}} \right)^2}$
${{\text{K}}_{{\text{sp}}}}\, = \,4{{\text{S}}^3}\, + \,0.40\,{{\text{S}}^2}$
As the ${\text{Ksp}}$ of lead iodide ${\text{Pb}}{{\text{I}}_{\text{2}}}$ is very-very ($1.4\, \times {10^{ - 8}}$) means the solubility of the lead iodide is very small so, we can neglect the ${{\text{S}}^3}$. On neglecting $4{{\text{S}}^3}$,
${{\text{K}}_{{\text{sp}}}}\, = \,0.40\,{{\text{S}}^2}$
Rearrange the equation for solubility as follows:
$\Rightarrow \,{{\text{S}}^2} = \,\,\dfrac{{{{\text{K}}_{{\text{sp}}}}}}{{0.40}}$
$\Rightarrow \,{\text{S}} = \,\,\sqrt {\dfrac{{{{\text{K}}_{{\text{sp}}}}}}{{0.40}}} $
Substitute $1.4\, \times {10^{ - 8}}$ for ${{\text{K}}_{{\text{sp}}}}$.
$\Rightarrow \,{\text{S}} = \,\,\sqrt {\dfrac{{1.4\, \times {{10}^{ - 8}}}}{{0.40}}} $
$\Rightarrow \,{\text{S}} = \,\,1.87\, \times {10^{ - 4}}$
So, the solubility of ${\text{Pb}}{{\text{I}}_{\text{2}}}$ in ${\text{Pb}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$ is $1.87\, \times {10^{ - 4}}$.

The compound sodium iodide ${\text{NaI}}$ dissociates as follows:
$\mathop {{\text{NaI}}}\limits_{0.010} \, \rightleftarrows \,\mathop {\,{\text{N}}{{\text{a}}^{\text{ + }}}}\limits_{0.010} {\text{ + }}\,\,\mathop {{{\text{I}}^ - }}\limits_{0.010} $
Sodium iodide produces iodide ion so the concentration of iodide ion is $0.010$M.
The solubility product for sodium iodide is represented as follows:
$\Rightarrow {{\text{K}}_{{\text{sp}}}}\, = \,{\text{P}}{{\text{b}}^{{\text{2 + }}}}{\text{ + }}\,\,2{{\text{I}}^ - }$
\[\Rightarrow {{\text{K}}_{{\text{sp}}}}\, = \,\left( {\text{S}} \right)\, \times \,\,{\left( {2\,{\text{S + }}\,{\text{0}}{\text{.010}}} \right)^2}\]
\[\Rightarrow {{\text{K}}_{{\text{sp}}}}\, = \,4{{\text{S}}^3}\, + \,0.040\,{{\text{S}}^2}\,{\text{ + }}\,\,1 \times {10^{ - 4}}{\text{S}}\]
On neglecting $4{{\text{S}}^3}$ and \[0.040\,{{\text{S}}^2}\],
$\Rightarrow {{\text{K}}_{{\text{sp}}}}\, = \,1 \times {10^{ - 4}}{\text{S}}$
Rearrange the equation for solubility as follows:
$\Rightarrow \,{\text{S}} = \,\,\dfrac{{{{\text{K}}_{{\text{sp}}}}}}{{1 \times {{10}^{ - 4}}}}$
Substitute $1.4\, \times {10^{ - 8}}$ for ${{\text{K}}_{{\text{sp}}}}$.
$\Rightarrow \,{\text{S}} = \,\,\dfrac{{1.4\, \times {{10}^{ - 8}}}}{{1 \times {{10}^{ - 4}}}}$
$\,{\text{S}} = \,\,1.4\, \times {10^{ - 4}}$

So, the solubility of ${\text{Pb}}{{\text{I}}_{\text{2}}}$ in ${\text{NaI}}$ is $1.4\, \times {10^{ - 4}}$.

Note: The formula of solubility product constant depends upon the number of ions produced by the ionic compounds in the water. When a soluble salt is dissolved in a solution and the soluble salt produces the ions that are produced by the insoluble salt that is already present in the solution, is known as the common ion effect. The common ion effect increases the solubility of the insoluble salt in the solution. Lead nitrates and sodium iodide both are soluble in water and they produce lead and iodide common ions respectively. So, the solubility of lead iodide is more in lead nitrates and sodium iodide than in water.