Question

The solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) in a \(7.2 \times 10^{-2}-M \mathrm{KIO}_{3}\) solution is \(6.0 \times 10^{-9} \mathrm{~mol} / \mathrm{L}\). Calculate the \(K_{\text {sp }}\) value 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 approximately \(1.87 \times 10^{-17}\).

Step-by-step solution

Write the dissociation equation

For the dissociation of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\), the equation is: \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s) \longleftrightarrow \mathrm{Pb}^{2+}(aq) + 2\mathrm{IO}_{3}^{-}(aq)\)

Find equilibrium concentrations of ions

According to the given data, the solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) is \(6.0 \times 10^{-9}\,\mathrm{mol\,L^{-1}}\). After dissociation: - \(\left[\mathrm{Pb^{2+}}\right]= 6.0 \times 10^{-9} \,\mathrm{M}\) - \(\left[\mathrm{IO_{3}^{-}}\right] = 7.2 \times 10^{-2}\,\mathrm{M} + 2(6.0 \times 10^{-9})\,\mathrm{M}\) As the second term in the above expression for \(\left[\mathrm{IO_{3}^{-}}\right]\) is much smaller than the first term, we can assume: \(\left[\mathrm{IO_{3}^{-}}\right] \approx 7.2 \times 10^{-2} \,\mathrm{M}\)

Calculate the solubility product constant

Using the equilibrium concentrations of the ions derived in step 2, we can calculate the \(K_{sp}\) for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\): \(K_{sp} = \left[\mathrm{Pb^{2+}}\right]\left[\mathrm{IO_{3}^-}\right]^2\) \(K_{sp} = (6.0 \times 10^{-9})(7.2 \times 10^{-2})^2\) \(K_{sp} = 1.87 \times 10^{-17}\) Therefore, the solubility product constant, \(K_{sp}\), for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) is approximately \(1.87 \times 10^{-17}\).

Key concepts

Dissociation Equation

Understanding how ionic solids dissolve in water is key to grasping solubility concepts. The process begins with a dissociation equation, which represents how a solid compound separates into its individual ions in a solution. For example, the dissociation of lead(II) iodate, written as \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\), can be illustrated by the equation:

\[\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s) \longleftrightarrow \mathrm{Pb}^{2+}(aq) + 2\mathrm{IO}_{3}^{-}(aq)\]
This reaction is reversible, indicating an equilibrium between the solid compound and its ions in solution. The right side shows the products of this dissociation: one lead ion (\(\mathrm{Pb}^{2+}\)) and two iodate ions (\(\mathrm{IO}_{3}^{-}\)). The coefficients signify the mole ratio in which they dissociate. It's these ratios that will later define the solubility product constant (\(K_{sp}\)).

Equilibrium Concentrations

At equilibrium, the concentrations of the ions in solution become constant. These equilibrium concentrations are crucial when computing solubility and the solubility product constant. For instance, if \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) has a solubility of \(6.0 \times 10^{-9}\) mol/L, then the concentration of lead ions in solution is \(6.0 \times 10^{-9} \mathrm{M}\).

Similarly, the concentration of iodate ions would supposedly double due to the stoichiometry of the dissociation equation. However, if \(\mathrm{IO}_{3}^{-}\) already exists in the solution from another source, such as \(\mathrm{KIO}_{3}\), its concentration will be significantly higher, and the contribution from \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) dissociation becomes negligible. Therefore, the equilibrium concentration of \(\mathrm{IO}_{3}^{-}\) can be approximated by the initial concentration of \(\mathrm{KIO}_{3}\), which in our example, is \(7.2 \times 10^{-2} \mathrm{M}\). This illustration shows the importance of considering all sources of ions when calculating equilibrium concentrations.

Solubility Equilibrium

Solubility equilibrium is the point at which the rate of dissolution and the rate of precipitation of a solute are equal, leading to a constant concentration of dissolved ions in solution. The numerically expressed representation of this equilibrium state for slightly soluble ionic compounds is known as the solubility product constant, \(K_{sp}\), which is particular to each substance at a given temperature.

The calculation of the \(K_{sp}\) involves multiplying the molarity of dissolved ions, each raised to the power of their respective coefficients in the balanced dissociation equation. With our compound \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\), the calculation would look like this:

\[K_{sp} = \left[\mathrm{Pb}^{2+}\right]\left[\mathrm{IO}_{3}^-\right]^2\]
By plugging in the equilibrium concentrations, we obtain the solubility product constant value. This constant helps us predict whether a precipitate will form under certain conditions and is instrumental for chemists to understand the solubility behavior of different compounds in various environments.