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Chapter 17: Problem 101

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

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

Step by step solution

01

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.
02

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.
03

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: \[ K_{\mathrm{sp}} = (2.4 \times 10^{-11}) \times (0.10)^2 = 2.4 \times 10^{-13} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

The maximum allowable concentration of \(\mathrm{Pb}^{2+}\) ions in drinking water is 0.05 ppm (i.e., \(0.05 \mathrm{~g}\) of \(\mathrm{Pb}^{2+}\) in 1 million grams of water). Is this guideline exceeded if an underground water supply is at equilibrium with the mineral anglesite \(\left(\mathrm{PbSO}_{4}\right)\left(K_{\mathrm{sp}}=1.6 \times 10^{-8}\right) ?\)

Acid-base reactions usually go to completion. Confirm this statement by calculating the equilibrium constant for each of the following cases: (a) A strong acid reacting with a strong base. (b) A strong acid reacting with a weak base \(\left(\mathrm{NH}_{3}\right) .(\mathrm{c})\) A weak acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) reacting with a strong base. (d) A weak acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) reacting with a weak base \(\left(\mathrm{NH}_{3}\right)\). (Hint: Strong acids exist as \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions and strong bases exist as \(\mathrm{OH}^{-}\) ions in solution. You need to look up \(K_{\mathrm{a}}, K_{\mathrm{b}}\), and \(K_{\mathrm{w}}\).)

A student carried out an acid-base titration by adding \(\mathrm{NaOH}\) solution from a burette to an Erlenmeyer flask containing an HCl solution and using phenolphthalein as the indicator. At the equivalence point, she observed a faint reddish-pink color. However, after a few minutes, the solution gradually turned colorless. What do you suppose happened?

Give an example to illustrate the general effect of complex ion formation on solubility.

Oil paintings containing lead(II) compounds as constituents of their pigments darken over the years. Suggest a chemical reason for the color change.
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