Question

The solubility of $\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)$ in a $0.10-M{\mathrm{KIO}}_3$ solution is $2.6\times {10}^{-11}\mathrm{mol}/\mathrm{L}.$ Calculate $K_{\mathrm{sp}}$ for $\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)$.

Short answer

The solubility product constant, $K_{sp}$, for $\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)$ is $6.76\times {10}^{-14}$.

Step-by-step solution

Write the chemical equation for the dissolution of $\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)$

The dissolution of $\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)$ in water can be represented as follows: $$\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)\rightleftharpoons {\mathrm{Pb}}^{2+}(aq)+2{\mathrm{IO}}_3^{-}(aq)$$

Set up the $K_{sp}$ expression

Now we can set up the solubility product constant expression based on the chemical equation. $K_{sp}$ is the product of the equilibrium concentrations of the ions, raised to the power of their stoichiometric coefficients. For the dissolution of $\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)$, the $K_{sp}$ expression is: $$K_{sp}=[{\mathrm{Pb}}^{2+}][{\mathrm{IO}}_3^{-}{]}^2$$

Determine the equilibrium concentrations of species involved

We are given that the solubility of $\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)$ in a $0.10-M{\mathrm{KIO}}_3$ solution is $2.6\times {10}^{-11}\mathrm{mol}/\mathrm{L}$. This means that the concentration of ${\mathrm{Pb}}^{2+}$, formed from the dissolution of $\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)$, is $2.6\times {10}^{-11}\mathrm{mol}/\mathrm{L}$. We are also given that the initial concentration of ${\mathrm{IO}}_3^{-}$ due to the presence of ${\mathrm{KIO}}_3$ is $0.10-M$. Since each mole of dissolved $\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)$ produces two moles of ${\mathrm{IO}}_3^{-}$, the equilibrium concentration of ${\mathrm{IO}}_3^{-}$ will be $0.10+2\times (2.6\times {10}^{-11})=0.10\mathrm{M}$ (Note: the added concentration is negligible since it is much smaller compared to $0.10-M$).

Substitute equilibrium concentrations into the $K_{sp}$ expression and solve

Now, we have all the required information. Substituting the equilibrium concentrations back into the $K_{sp}$ expression, we get: $$K_{sp}=(2.6\times {10}^{-11})(0.10{)}^2$$ Calculate $K_{sp}$: $$K_{sp}=6.76\times {10}^{-14}$$ Thus, the solubility product constant, $K_{sp}$, for $\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)$ is $6.76\times {10}^{-14}$.

Key concepts

Solubility Equilibrium

Solubility equilibrium refers to a dynamic balance between the opposing processes of dissolution and precipitation of a solute in a solvent, typically water. At this equilibrium, the rate at which the solute dissolves in the solvent is equal to the rate at which it precipitates back into solid form. This state can be disturbed by changing the temperature, pressure, or the presence of other substances that might interact with the solute or solvent.

In the given exercise, we have a sparingly soluble compound, $\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)$, which reaches solubility equilibrium in a $0.10-M{\mathrm{KIO}}_3$ solution. Its solubility equilibrium can be represented by the equation:
$$\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)\rightleftharpoons {\mathrm{Pb}}^{2+}(aq)+2{\mathrm{IO}}_3^{-}(aq)$$.
The concentration of dissolved ${\mathrm{Pb}}^{2+}$ ions at equilibrium conveys the extent to which $\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)$ can dissolve under the provided conditions.

Equilibrium Constant Expression

The equilibrium constant expression for a solubility equilibrium is known as the solubility product constant, denoted by $K_{sp}$. It is a quantitative measure of the solubility of a compound under a given set of conditions and is determined by the concentrations of the ions in solution at equilibrium raised to the power of their coefficients in the balanced dissolution equation.

For the dissolution process $$\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)\rightleftharpoons {\mathrm{Pb}}^{2+}(aq)+2{\mathrm{IO}}_3^{-}(aq)$$, the $K_{sp}$ expression is:$$K_{sp}=[{\mathrm{Pb}}^{2+}][{\mathrm{IO}}_3^{-}{]}^2$$.
In this case, calculating the $K_{sp}$ helps us understand the solubility of $\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)$ in the presence of a common ion, which in this scenario is ${\mathrm{IO}}_3^{-}$ from ${\mathrm{KIO}}_3$.

Chemical Dissolution

Chemical dissolution is the process through which a solid substance (solute) disperses into individual ions or molecules in a solvent, forming a solution. This process can be influenced by factors such as temperature, pressure, and the nature of the solute and solvent. The ease of dissolution is determined by the solute's solubility in the particular solvent.

In our exercise, the chemical dissolution of $\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)$ in water involves breaking apart of the lattice structure of the solid into its constituent ions, ${\mathrm{Pb}}^{2+}$ and ${\mathrm{IO}}_3^{-}$, which then become surrounded by water molecules in a process known as hydration. This particular dissolution is limited due to low product solubility, as evidenced by the extremely low solubility value $2.6\times {10}^{-11}\mathrm{mol}/\mathrm{L}$.

Stoichiometry in Equilibrium Reactions

Stoichiometry in equilibrium reactions is the quantitative relationship between the reactants and products in a fully balanced chemical equation at equilibrium. It is essential for calculating the equilibrium constant as well as determining the relationship between the amounts of substances involved in the reaction.

In this solubility context, stoichiometry helps us understand that for each mole of $\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2(s)$ that dissolves, it generates one mole of ${\mathrm{Pb}}^{2+}$ and two moles of ${\mathrm{IO}}_3^{-}$. With the stoichiometry of the dissolution reaction, we can calculate how the presence of excess ${\mathrm{IO}}_3^{-}$ from ${\mathrm{KIO}}_3$ affects the equilibrium by shifting it, in accordance with Le Chatelier's principle, which ultimately impacts the solubility and the $K_{sp}$ of the compound.