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Chapter 16: Problem 107

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
The solubility product constant, \(K_{sp}\), for \(\mathrm{Pb(IO_3)_2}\) is \(2.4 \times 10^{-13}\)

Step by step solution

01

Establish the Dissociation Reaction of \(\mathrm{Pb(IO_{3})_{2}}\)

Consider the dissociation reaction of \(\mathrm{Pb(IO_{3})_{2}}\). It can be written as: \[\mathrm{Pb(IO_{3})_{2}}(s) \leftrightarrow \mathrm{Pb^{2+}}(aq) + 2\mathrm{IO_{3}^{-}}(aq)\]
02

Write down initial concentrations and changes

In a solution of \(0.10 \mathrm{M} \mathrm{NaIO_{3}}\), the equilibrium concentration of \(\mathrm{IO_{3}^-}\) is \(0.10 \mathrm{M}\) due to the presence of \(\mathrm{NaIO_{3}}\). Since \(\mathrm{Pb(IO_{3})_{2}}\) is sparingly soluble, a very small amount of it dissolves to give \(\mathrm{Pb^{2+}}\) ions. This amount is the molar solubility of the compound, \(2.4 \times 10^{-11} \mathrm{mol/L}\). So, \[Pb^{2+} = 2.4 \times 10^{-11} \mathrm{mol/L}\].
03

Find the concentration of IO3-

The dissolving of one \(\mathrm{Pb(IO_{3})_{2}}\) molecule produces two \(\mathrm{IO_{3}^{-}}\) ions. Therefore, the change in concentration of \(\mathrm{IO_{3}^{-}}\) is \(2 \times \text{molar solubility of} \, \mathrm{Pb(IO_{3})_{2}}\). So, \[\mathrm{IO_{3}^{-}} = 0.10 + 2 \times 2.4 \times 10^{-11} \approx 0.10M\]. The concentration does not significantly change as the solubility of \(\mathrm{Pb(IO_{3})_{2}}\) is extremely low.
04

Calculate \(K_{sp}\) using the equilibrium concentrations

By substituting the equilibrium concentrations of \(\mathrm{Pb^{2+}}\) and \(\mathrm{IO_{3}^{-}}\) into the \(K_{sp}\) expression, this gives: \[K_{sp} = [Pb^{2+}][IO_3^-]^2 = (2.4 \times 10^{-11})(0.10)^2 = 2.4 \times 10^{-13}\]

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