Question

The K_sp for lead iodide is 1.4 x 10^-8. Calculate the solubility of lead iodide in each of the following.

1. Water
2. 0.10 MPb(NO₃)₂
3. 0.10 MNal

Short answer

1. The solubility of lead iodide in water is x = 1.5 x 10^-3M
2. The solubility of lead iodide in 0.10 MPb(NO₃)₂ is x = 1.9 x 10^-4M
3. The solubility of lead iodide in 0.10 MNal is x = 1.4 x 10^-4M

Step-by-step solution

Definition of solubility

The maximum amount of a substance that will dissolve in a given amount of solvent at a specified temperature is defined as solubility.

Calculation of solubility

a. $K_{sp}\left(Pb{I}_2\right)=1.4\times {10}^{-8}$

The dissociation reaction for lead iodide is

$Pb{I}_2\left(s\right)\rightleftharpoons P{b}^{2+}\left(aq\right)+2I^{-}\left(aq\right)$

The expression for the solubility constant is

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

Let the equilibrium concentrations be:

Concentration(mol/L)	$Pb{I}_2\rightleftharpoons$	$P{b}^{2+}$+	2I
Initial	x	0	0
Change	-x	+x	+2x
Equilibrium	0	x	2x

Substituting the values as:

localid="1649137189929" $$\begin{gathered} \mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{=}\left[x\right]{\left[2x\right]}^{\mathbf{2}} \\ \mathbf{4}{\mathit{x}}^{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}} \\ {\mathit{x}}^{\mathbf{3}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}} \\ \mathit{x}\mathbf{=}\sqrt[\mathbf{3}]{\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}} \\ \mathit{x}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{M} \end{gathered}$$

b.

The dissociation reaction for lead iodide is

$Pb{I}_2\left(s\right)\rightleftharpoons P{b}^{2+}\left(aq\right)+2I^{-}\left(aq\right)$

The expression for the solubility constant is

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

Let the equilibrium concentrations be:

Concentration(mol/L)	$Pb{I}_2\rightleftharpoons$	$P{b}^{2+}$+	2I
Initial	x	0.10	0
Change	-x	+x	+2x
Equilibrium	0	0.10+x	2x

Substituting the values as:

localid="1649137339740" $$\begin{gathered} \mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{=}\left[0.10+x\right]{\left[2x\right]}^{\mathbf{2}} \\ \mathit{x}\mathbf{<}\mathbf{<}\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}\mathit{s}\mathit{o} \\ \mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{=}\left[0.10+x\right]{\left[2x\right]}^{\mathbf{2}} \\ {\mathit{x}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}}{\mathbf{0}\mathbf{.}\mathbf{4}} \\ \mathit{x}\mathbf{=}\sqrt{\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}} \\ \mathit{x}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{M} \end{gathered}$$

c.

The dissociation reaction for lead iodide is

$Pb{I}_2\left(s\right)\rightleftharpoons P{b}^{2+}\left(aq\right)+2I^{-}\left(aq\right)$

The expression for the solubility constant is

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

Let the equilibrium concentrations be:

Concentration(mol/L)	$Pb{I}_2\rightleftharpoons$	$P{b}^{2+}$+	2I
Initial	x	0	0.01
Change	-x	+x	+2x
Equilibrium	0	x	0.010+2x

Substituting the values as:

role="math" localid="1649137505787" $$\begin{gathered} \mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{=}\left[x\right]{\left[0.010+2x\right]}^{\mathbf{2}} \\ \mathit{x}\mathbf{<}\mathbf{<}\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{,} \\ \mathit{x}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}}{\mathbf{0}\mathbf{.}\mathbf{0001}} \\ \mathit{x}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{M} \end{gathered}$$