Question

A saturated solution of lead(II) iodide, PbI2 has an iodide concentration of 3.0 x 10^-3 mol/L.

a) What is the molar solubility of PbI2?

b) Determine the solubility constant, Ksp, for lead(II) iodide.

c) Does the molar solubility of lead (II) iodide increase, decrease, or remain unchanged with the addition of potassium iodide to the solution? EXPLAIN.

Answer 1

Concepts and reason

The concepts used to solve this question are to determine the molar solubility, solubility product ${K_{sp}}$ of lead iodide and effect of the molar solubility with the addition of potassium iodide.

Fundamentals

Molar solubility is the amount of substance (moles) that can be dissolved in one liter solution.

Multiplying the molar solubility with the molar mass gives the solubility in ${\rm{g/L}}$ .

The solubility product, ${K_{sp}}$ is the equilibrium constant for a solid that is dissolved in a solution. It is the product of the concentration of the products raised to the power of their coefficients.

According to Le Chatelier’s principle, if equilibrium of a system is disturbed, the position of the equilibrium shifts so as to counteract that change.

Here, with the addition of potassium iodide, iodide ion concentration increases. So, the equilibrium will shift so that the iodide concentration decreases. More ${\rm{Pb}}{{\rm{I}}_2}$ is formed.

The dissociation of ${\rm{Pb}}{{\rm{I}}_2}$ is as follows:

${\rm{Pb}}{{\rm{I}}_2}\longrightarrow{{}}{\rm{P}}{{\rm{b}}^{2 + }} + 2{{\rm{I}}^ - }$

The expression for the solubility product for the reaction is as follows:

${K_{sp}} = \left[ {{\rm{P}}{{\rm{b}}^{2 + }}} \right]{\left[ {{{\rm{I}}^ - }} \right]^2}$

Let x be the molar solubility.

Given,

$\left[ {{{\rm{I}}^ - }} \right] = 3.0 \times {10^{ - 3}}{\rm{ mol/L}}$

Construct and ICE table for the reaction as follows:

$\begin{array}{l}\\{\rm{ Pb}}{{\rm{I}}_2}\longrightarrow{{}}{\rm{P}}{{\rm{b}}^{2 + }} + 2{{\rm{I}}^ - }\\\\{\rm{Initial: 0 0}}\\\\{\rm{Change: }} + x{\rm{ }} + 2x\\\\{\rm{Equilibrium: }}x{\rm{ }}2x\\\end{array}$

$\begin{array}{c}\\\left[ {{{\rm{I}}^ - }} \right] = 2x\\\\{\rm{ }} = 3.0 \times {10^{ - 3}}{\rm{ mol/L}}\\\\x = \frac{{3.0 \times {{10}^{ - 3}}{\rm{ mol/L}}}}{2}\\\\{\rm{ = }}1.5 \times {10^{ - 3}}{\rm{ mol/L}}\\\\\left[ {{\rm{P}}{{\rm{b}}^{2 + }}} \right] = x\\\\{\rm{ }} = 1.5 \times {10^{ - 3}}{\rm{ mol/L}}\\\end{array}$

The expression for the solubility product for the reaction is as follows:

${K_{sp}} = \left[ {{\rm{P}}{{\rm{b}}^{2 + }}} \right]{\left[ {{{\rm{I}}^ - }} \right]^2}$

Given ,

$\left[ {{{\rm{I}}^ - }} \right] = 3.0 \times {10^{ - 3}}{\rm{ mol/L}}$

$\left[ {{\rm{P}}{{\rm{b}}^{2 + }}} \right] = 1.5 \times {10^{ - 3}}{\rm{ mol/L}}$

The solubility product constant is calculated as follows:

$\begin{array}{l}\\{K_{sp}} = \left( {1.5 \times {{10}^{ - 3}}{\rm{ mol/L}}} \right){\left( {3.0 \times {{10}^{ - 3}}{\rm{ mol/L}}} \right)^2}\\\\{\rm{ = 1}}{\rm{.35}} \times {\rm{1}}{{\rm{0}}^{ - 8}}\\\end{array}$

The dissociation of ${\rm{Pb}}{{\rm{I}}_2}$ is as follows:

${\rm{Pb}}{{\rm{I}}_2}\longrightarrow{{}}{\rm{P}}{{\rm{b}}^{2 + }} + 2{{\rm{I}}^ - }$

The expression for the solubility product for the reaction is as follows:

${K_{sp}} = \left[ {{\rm{P}}{{\rm{b}}^{2 + }}} \right]{\left[ {{{\rm{I}}^ - }} \right]^2}$

Potassium iodide splits into ${{\rm{K}}^ + }{\rm{ and }}{{\rm{I}}^ - }$ .

With the addition of potassium iodide, the molar solubility of ${\rm{Pb}}{{\rm{I}}_2}$ decreases.

Ans:

Therefore, the molar solubility is $1.5 \times {10^{ - 3}}{\rm{ mol/L}}$ .