Question

The molar solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}\) in a \(0.10 \mathrm{M} \mathrm{NaIO}_{3}\) solution is \(2.4 \times 10^{-11} \mathrm{~mol} / \mathrm{L}\). What is \(K_{\mathrm{sp}}\) for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}\) ?

Short answer

\( K_{\mathrm{sp}} = 2.4 \times 10^{-13} \)

Step-by-step solution

Understand the Dissolution Equation

The dissolution of lead(II) iodate, \( \mathrm{Pb(IO_3)_2} \), in water is represented by the equation: \[ \mathrm{Pb(IO_3)_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2 IO_3^- (aq)} \] This shows that one mole of \( \mathrm{Pb(IO_3)_2} \) dissolves to give one mole of \( \mathrm{Pb^{2+}} \) ions and two moles of \( \mathrm{IO_3^-} \) ions.

Relate Molar Solubility to Ion Concentrations

Given the molar solubility of \( \mathrm{Pb(IO_3)_2} \) is \( 2.4 \times 10^{-11} \) M, the concentration of \( \mathrm{Pb^{2+}} \) ions is the same as the molar solubility. The concentration of \( \mathrm{IO_3^-} \) ions is twice the molar solubility plus the \( \mathrm{NaIO_3} \) concentration. Thus: \[ [\mathrm{Pb^{2+}}] = 2.4 \times 10^{-11} \text{ M} \] \[ [\mathrm{IO_3^-}] = 2 \times 2.4 \times 10^{-11} + 0.10 = 0.10 \text{ M} \] because the added \( \mathrm{NaIO_3} \) contributes to the \( \mathrm{IO_3^-} \) concentration.

Write and Solve for Ksp

The solubility product \( K_{\mathrm{sp}} \) expression for \( \mathrm{Pb(IO_3)_2} \) is given by: \[ K_{\mathrm{sp}} = [\mathrm{Pb^{2+}}] [\mathrm{IO_3^-}]^2 \] Substitute the values found earlier: \[ K_{\mathrm{sp}} = (2.4 \times 10^{-11}) \times (0.10)^2 = 2.4 \times 10^{-13} \]

Key concepts

Molar Solubility

Molar solubility is a key concept in understanding how much of a compound can dissolve in a solution before it becomes saturated. It refers to the number of moles of a solute that can dissolve per liter of solution. In our problem, we are examining the molar solubility of lead(II) iodate, \( \mathrm{Pb(IO_3)_2} \), which is given as \( 2.4 \times 10^{-11} \mathrm{~mol/L} \).
This tiny value indicates that lead(II) iodate is only sparingly soluble, meaning it does not dissolve well in water. When calculating molar solubility:

• We express it in units of mol/L.
• It tells us directly about the concentration of dissolved \( \mathrm{Pb^{2+}} \) ions in the solution.

Knowing the molar solubility helps calculate other important concentrations and find the solubility product constant, \( K_{\mathrm{sp}} \).

Dissolution Equation

The dissolution equation is a chemical equation that shows how a compound dissociates into its ions in a solution.For lead(II) iodate, the equation is given by:
\[ \mathrm{Pb(IO_3)_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2 IO_3^- (aq)} \]
This equation shows that when \( \mathrm{Pb(IO_3)_2} \) dissolves, it separates into one \( \mathrm{Pb^{2+}} \) ion and two \( \mathrm{IO_3^-} \) ions. Key points:

• The equation helps us understand the stoichiometry of the dissolution process.
• It shows the relationship between the different ions produced in the solution.

By studying this equation, we can understand how molar solubility affects ion concentrations in a solution.

Ion Concentration

Ion concentration refers to the amount of each ion present in the solution due to the dissolution of lead(II) iodate. Since the molar solubility of \( \mathrm{Pb(IO_3)_2} \) is \( 2.4 \times 10^{-11} \mathrm{M} \), this is also the concentration of \( \mathrm{Pb^{2+}} \) ions in the solution.
For \( \mathrm{IO_3^-} \) ions, the concentration is affected by both the dissolution and the presence of \( \mathrm{NaIO_3} \):

• Each mole of \( \mathrm{Pb(IO_3)_2} \) contributes two moles of \( \mathrm{IO_3^-} \) ions.
• So, their concentration is \( 2 \times 2.4 \times 10^{-11} + 0.10 \), which results in \( 0.10 \mathrm{M} \) due to the dominant \( \mathrm{NaIO_3} \).

Understanding ion concentrations helps in calculating the \( K_{\mathrm{sp}} \), which reflects the saturation level in the solution.

Lead(II) Iodate

Lead(II) iodate, \( \mathrm{Pb(IO_3)_2} \), is a sparingly soluble ionic compound. When it dissolves, it provides important insights into solubility equilibria, specifically in the calculation of the solubility product constant, \( K_{\mathrm{sp}} \).
Here's why lead(II) iodate is significant in this context:

• Understanding its solubility characteristics allows chemists to predict the compound's behavior in different solutions.
• Its dissolution involves a well-defined stoichiometric relationship, simplifying solubility calculations.

By knowing the solubility and dissolution behavior of \( \mathrm{Pb(IO_3)_2} \), one can deduce crucial information about solutions, which is helpful in various chemical applications.