Chapter 17, Problem 63QP

The solubility product of ${\text{PbBr}}_{\text{2}}\text{is 8}\text{.9 }\times {\text{ 10}}^{\text{-6}}$. Determine the molar solubility in (a) pure water, (b) 0.20 M KBr solution, and (c) $\text{0}\text{.20}M\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ solution.

Interpretation Introduction

Interpretation:

The molar solubility of $\text{Pb}{\left(\text{Br}\right)}_{\text{2}}$ in pure water, in $\text{KBr}$ solution, and in $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ solutions are to be determined with given $K_{\text{sp}}$ value of $\text{Pb}{\left(\text{Br}\right)}_{\text{2}}$.

Concept introduction:

The amount of solute dissolved in a given volume of the solvent to form a saturated solution at a given temperature is termed as solubility of the solute in the solvent at that temperature.

The solubility product of the sparingly soluble salt is given as the product of the concentration of the ions raised to the power equal to the number of times the ion occurs in the equation, after dissociation of the electrolyte.

Number of moles of solute dissolved per litre of solution is called molar solubility.

At a given temperature, the product of molar concentrations of the ions of a salt present in the solution is known as the solubility product of the salt. It is represented by $K_{sp}$.

Higher is the value of solubility product of a salt, higher is its solubility.

The presence of common ions in the solution decreases the solubility of a given compound.

For a general reaction: $\text{AB}\left(s\right)\rightleftharpoons {\text{A}}^{+}\left(aq\right)+{\text{B}}^{-}\left(aq\right)$

The solubility product can be calculated by the expression as: ${\text{K}}_{\text{sp}}{\text{= [A}}^{\text{+}}{\text{][B}}^{-}\text{]}$

Here, $K_{\text{sp}}$ is solubility product constant and $sp$ stands for solubility product.

The molar solubility of a compound is directly proportional to the number of molecules present in the given amount of solvent.

Answer to Problem 63QP

Solution:

The molar solubility of $\text{Pb}{\left(\text{Br}\right)}_{\text{2}}$ in pure water is $1.3\times {10}^{-2}\text{M}$.

The molar solubility of $\text{Pb}{\left(\text{Br}\right)}_{\text{2}}$ in $\text{KBr}$ solution is $2.2\times {10}^{-4}\text{M}$.

The molar solubility of $\text{Pb}{\left(\text{Br}\right)}_{\text{2}}$ in $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ solution is $3.3\times {10}^{-3}\text{M}$,

Explanation of Solution

a) Pure water

Thesolubility product constant of $\text{Pb}{\left(\text{Br}\right)}_{\text{2}}$ is $8.9\times {10}^{-6}\text{M}$,

The equation of the dissociation of $\text{Pb}{\left(\text{Br}\right)}_{\text{2}}$ in the aqueous solution is as follows:

$\text{Pb}{\left(\text{Br}\right)}_{\text{2}}\left(s\right)\rightleftharpoons {\text{Pb}}^{\text{2+}}\left(aq\right){\text{+ Br}}^{-}\left(aq\right)$

Consider s to be the molar solubility.

The molar solubility of $\text{Pb}{\left(\text{Br}\right)}_{\text{2}}$ in pure water is calculated as follows:

Summarize the concentration at the equilibrium as follows:

$\begin{array}{l}\text{Pb}{\left(\text{Br}\right)}_2\left(s\right)\rightleftharpoons {\text{Pb}}^{2+}\left(aq\right)+2{\text{Br}}^{-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& -& 0& 0\\ \text{Change}\left(\text{M}\right)& -s& +s& +2s\\ \text{Equilibrium}\left(\text{M}\right)& -& s& 2s\end{array}\end{array}$

The equilibrium expression for a reaction is written as follows:

$K_{\text{sp}}=[{\text{Pb}}^{2+}]{[{\text{Br}}^{-}]}^2$

Here, $K_{\text{sp}}$ is solubility product constant, ${\text{Pb}}^{2+}$ is the concentration of lead ion, and ${\text{Br}}^{-}$ ion is the concentration bromine ion.

Substitute the value of $K_{\text{sp}}$, ${\text{Pb}}^{2+}$, and ${\text{Br}}^{-}$ ion in the above expression,

$\begin{array}{c}{\text{K}}_{\text{sp}}=\left(s\right){\left(2s\right)}^2\\ 8.9\times {10}^{-6}=4s^3\\ s={\left(\frac{8.9\times {10}^{-6}}{4}\right)}^{\frac{1}{3}}\\ s=1.3\times {10}^{-2}\text{M}\end{array}$

Hence, the molar solubility of $\text{Pb}{\left(\text{Br}\right)}_{\text{2}}$ in water is $\text{1}\text{.3}\times {\text{10}}^{-2}\text{M}$.

b) $\text{0}\text{.20}\text{M}$ $\text{KBr}$

The solubility product constant of $\text{Pb}{\left(\text{Br}\right)}_{\text{2}}$ is $8.9\times {10}^{-6}\text{M}$.

The molar solubility of $\text{Pb}{\left(\text{Br}\right)}_{\text{2}}$ in $\text{KBr}$ solution is calculated as follows:

The equation of the dissociation of $\text{KBr}$ is as follows:

$\text{KBr}\left(s\right)\rightleftharpoons {\text{K}}^{\text{+}}\left(aq\right)+{\text{ Br}}^{-}\left(aq\right)$

$\text{KBr}$ is the strong electrolyte. It completely dissociates into cation and anion in water. It is highlysoluble in an aqueous solution.

Summarize the concentration at the equilibrium as follows:

$\begin{array}{l}\text{KBr}\left(s\right)\to {\text{K}}^{+}\left(aq\right)+{\text{ Br}}^{-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& 0.20& 0& 0\\ \text{Change}\left(\text{M}\right)& -0.20& +0.20& +0.20\\ \text{Equilibrium}\left(\text{M}\right)& 0& 0.20& 0.20\end{array}\end{array}$

Therefore, the concentration of ${\text{[Br}}^{-}\text{]}$ is $0.20\text{M}$.

Summarize the concentration at the equilibrium as follows:

Consider s to be the molar solubility.

$\begin{array}{l}\text{Pb}{\left(\text{Br}\right)}_2\left(s\right)\rightleftharpoons {\text{Pb}}^{2+}\left(aq\right)+2{\text{Br}}^{-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& -& 0& 0.20\\ \text{Change}\left(\text{M}\right)& -s& +s& +2s\\ \text{Equilibrium}\left(\text{M}\right)& -& s& 0.20+2s\end{array}\end{array}$

The equilibrium expression for a reaction is written as:

$K_{\text{sp}}=[{\text{Pb}}^{2+}]{[{\text{Br}}^{-}]}^2$

Here, $K_{\text{sp}}$ is solubility product constant, ${\text{Pb}}^{2+}$ is the concentration of lead ion, and ${\text{Br}}^{-}$ ion is the concentration bromine ion.

Substitute the value of $K_{\text{sp}}$, ${\text{Pb}}^{2+}$, and ${\text{Br}}^{-}$ ion in the above expression.

$K_{\text{sp}}=\left(s\right){\left(2s+0.20\right)}^2$

The value of s is very small as compared to 0.20. It can be neglected.

$\begin{array}{l}8.9\times {10}^{-6}=s\times {\left(0.20\right)}^2\\ s=2.2\times {10}^{-4}\text{M}\end{array}$

Hence, the molar solubility of $\text{Pb}{\left(\text{Br}\right)}_{\text{2}}$ in $\text{KBr}$ solution is $2.2\times {10}^{-4}\text{M}$.

c) $\text{0}\text{.20}\text{M}$ $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$

The solubility product constant of $\text{Pb}{\left(\text{Br}\right)}_{\text{2}}$ is $8.9\times {10}^{-6}\text{M}$.

The molar solubility of $\text{Pb}{\left(\text{Br}\right)}_{\text{2}}$ in $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ solution is calculated as follows:

The equation of the dissociation of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$

$\text{Pb}{\left({\text{NO}}_3\right)}_2\left(s\right)\rightleftharpoons {\text{Pb}}^{\text{2+}}\left(aq\right){\text{+ 2NO}}_3{}^{-}\left(aq\right)$

$\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is the strong electrolyte. It completely dissociates into cation and anion in water. It is highly soluble in an aqueous solution.

Summarize the concentration at the equilibrium as follows:

$\begin{array}{l}\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(s\right)\to {\text{Pb}}^{2+}\left(aq\right)+2{\text{NO}}_3^{-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& 0.20& 0& 0\\ \text{Change}\left(\text{M}\right)& -0.20& +0.20& +2\times 0.20\\ \text{Equilibrium}\left(\text{M}\right)& 0& 0.20& 0.40\end{array}\end{array}$

The concentration of ${\text{[Pb}}^{\text{2+}}\text{]}$ is $\text{0}\text{.20}\text{M}$.

Summarize the concentration at the equilibrium as follows:

Consider s to be the molar solubility.

$\begin{array}{l}\text{Pb}{\left(\text{Br}\right)}_{\text{2}}\left(s\right)\rightleftharpoons {\text{Pb}}^{\text{2+}}\left(aq\right)+{\text{2Br}}^{-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& -& 0.20& 0\\ \text{Change}\left(\text{M}\right)& -s& +s& +2s\\ \text{Equilibrium}\left(\text{M}\right)& -& 0.20+s& 2s\end{array}\end{array}$

The equilibrium expression for a reaction is written as:

$K_{\text{sp}}=[{\text{Pb}}^{2+}]{[{\text{Br}}^{-}]}^2$

Here, $K_{\text{sp}}$ is solubility product constant, ${\text{Pb}}^{2+}$ is the concentration of lead ion, and ${\text{Br}}^{-}$ ion is the concentration bromine ion.

Substitute the value of $K_{\text{sp}}$, ${\text{Pb}}^{2+}$, and ${\text{Br}}^{-}$ ion in the above expression,

$K_{\text{sp}}=\left(s+0.20\right){\left(2s\right)}^2$

The value of s is very small as compared to 0.20. It can be neglected.

$\begin{array}{l}8.9\times {10}^{-6}=s^2\times \left(0.20\right)\times 4\\ s^2=\frac{8.9\times {10}^{-6}}{0.80}\\ s={\left(\frac{8.9\times {10}^{-6}}{0.80}\right)}^{\frac{1}{2}}\\ s=3.3\times {10}^{-3}\text{M}\end{array}$

Hence, the molar solubility of $\text{Pb}{\left(\text{Br}\right)}_{\text{2}}$ in $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ solution is $3.3\times {10}^{-3}\text{M}$.