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Chapter 16: Problem 95

The \(K_{\text {? }}\) for \(\mathrm{PbI}_{2}(s)\) is \(1.4 \times 10^{-8} .\) Calculate the solubility of \(\mathrm{PbI}_{2}(s)\) in \(0.048 M\) NaI.

Short Answer

Expert verified
The solubility of \(\mathrm{PbI_2}(s)\) in a \(0.048\, \text{M}\) NaI solution is approximately \(6.1 \times 10^{-6}\, \text{M}\).

Step by step solution

01

Identify the solubility product constant

The solubility product constant, K\(_{\text {sp}}\), for \(\mathrm{PbI}_{2}(s)\) is given as \(1.4 \times 10^{-8}\).
02

Write the balanced dissolution reaction and the corresponding solubility product expression

Write the balanced dissolution reaction for \(\mathrm{PbI}_{2}(s)\) and the corresponding solubility product expression for K\(_{\text {sp}}\): \[\mathrm{PbI}_{2}(s) \rightleftharpoons \mathrm{Pb^{2+}}(aq) + 2 \mathrm{I^{-}}(aq)\] \[K_{\text{sp}} = [\mathrm{Pb^{2+}}][\mathrm{I^{-}}]^2\]
03

Set up the reaction table

Use a table to represent the initial concentration of ions before dissolution, change in concentration during dissolution, and the equilibrium concentration of ions at the end of dissolution. Before dissolution: \[ [\mathrm{Pb^{2+}}] = 0\] \[ [\mathrm{I^{-}}] = 0.048\,\text{M}\, \text{(from NaI)}\] Change during dissolution: \[ [\mathrm{Pb^{2+}}] \rightarrow +s\] \[ [\mathrm{I^{-}}] \rightarrow +2s\] At equilibrium: \[ [\mathrm{Pb^{2+}}] = s\] \[ [\mathrm{I^{-}}] = 0.048 \, \text{M} + 2s\]
04

Substitute the equilibrium concentrations into the K\(_{\text {sp}}\) expression

Substitute the equilibrium concentrations into the expression for K\(_{\text {sp}}\): \[K_{\text{sp}}= 1.4 \times 10^{-8} = s (0.048 + 2s)^2\]
05

Solve for solubility 's'

Since K\(_{\text {sp}}\) is a small value, the value of 's' will be also small in comparison to 0.048. Therefore, we can approximate the equation by neglecting the 2s term: \[1.4 \times 10^{-8} = s (0.048)^2\] Now, solve for 's': \[s = \frac{1.4 \times 10^{-8}}{(0.048)^2} = 6.1 \times 10^{-6}\, \text{M}\]
06

Conclusion

The solubility of \(\mathrm{PbI_2}(s)\) in a \(0.048\, \text{M}\) NaI solution is approximately \(6.1 \times 10^{-6}\, \text{M}\).

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Most popular questions from this chapter

Calculate the equilibrium concentrations of \(\mathrm{NH}_{3}, \mathrm{Cu}^{2+}\), \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+}\), and \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}\) in a solution prepared by mixing \(500.0 \mathrm{~mL}\) of \(3.00 \mathrm{M} \mathrm{NH}_{3}\) with \(500.0 \mathrm{~mL}\) of \(2.00 \times 10^{-3} M \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\). The stepwise equilib- ria are \(\mathrm{Cu}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{CuNH}_{3}^{2+}(a q)\) \(K_{1}=1.86 \times 10^{4}\) \(\mathrm{CuNH}_{3}{ }^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}(a q)\) \(K_{2}=3.88 \times 10^{3}\) \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}{ }^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+}(a q)\) \(K_{3}=1.00 \times 10^{3}\) \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}{ }^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}(a q)\) \(K_{4}=1.55 \times 10^{2}\)

List some ways one can increase the solubility of a salt in water.

Which of the following will affect the total amount of solute that can dissolve in a given amount of solvent? a. The solution is stirred. b. The solute is ground to fine particles before dissolving. c. The temperature changes.

For which salt in each of the following groups will the solubility depend on \(\mathrm{pH}\) ? a. \(\mathrm{AgF}, \mathrm{AgCl}, \mathrm{AgBr}\) c. \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{Sr}\left(\mathrm{NO}_{2}\right)_{2}\) b. \(\mathrm{Pb}(\mathrm{OH})_{2}, \mathrm{PbCl}_{2}\) d. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{Ni}(\mathrm{CN})_{2}\)

Consider \(1.0 \mathrm{~L}\) of an aqueous solution that contains \(0.10 \mathrm{M}\) sulfuric acid to which \(0.30\) mole of barium nitrate is added. Assuming no change in volume of the solution, determine the \(\mathrm{pH}\), the concentration of barium ions in the final solution, and the mass of solid formed.
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