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Chapter 17: Problem 60

Consider a beaker containing a saturated solution of \(\mathrm{PbI}_{2}\) in equilibrium with undissolved \(\mathrm{PbI}_{2}(s)\). Now solid \(\mathrm{K} \mathrm{I}\) is added to this solution. (a) Will the amount of solid \(\mathrm{PbI}_{2}\) at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of \(\mathrm{Pb}^{2+}\) ions in solution increase or decrease? (c) Will the concentration of \(I\) ions in solution increase or decrease?

Short Answer

Expert verified
(a) The amount of solid \(\mathrm{PbI}_{2}\) at the bottom of the beaker will increase due to the common ion effect and Le Chatelier's principle, as the \(I^-\) concentration increases from the dissolution of \(\mathrm{KI}\). (b) The concentration of \(\mathrm{Pb}^{2+}\) ions in the solution will decrease as more \(\mathrm{PbI}_{2}\) precipitates. (c) The concentration of \(I^-\) ions in the solution will increase despite the precipitation of \(\mathrm{PbI}_{2}\), due to the dissociation of \(\mathrm{KI}\).

Step by step solution

01

Write the solubility equilibrium for \(\mathrm{PbI}_{2}\)

The solubility equilibrium for \(\mathrm{PbI}_{2}\) can be expressed as: \[\mathrm{PbI}_{2} \rightleftharpoons \mathrm{Pb}^{2+} + 2\mathrm{I}^-\] The \(K_{sp}\) expression for this equilibrium is: \[K_{sp} = [\mathrm{Pb}^{2+}] [I^-]^2\] Remember that for this particular situation, the solution is saturated.
02

Adding solid \(\mathrm{KI}\) to the beaker

When solid \(\mathrm{KI}\) is added to the saturated solution, it will dissolve and dissociate into \(\mathrm{K}^+\) and \(I^-\) ions. The equilibrium will be affected by the increase in concentration of \(I^-\) ions from the dissociation of \(\mathrm{KI}\). The reaction for the dissolution of \(\mathrm{KI}\) is: \[\mathrm{KI} \rightarrow \mathrm{K}^+ + I^-\] Since \(\mathrm{PbI}_{2}\) and \(\mathrm{KI}\) have a common ion (\(I^-\)), the common ion effect comes into play.
03

Analyzing the effect on the solubility of \(\mathrm{PbI}_{2}\)

The increase in the concentration of \(I^-\) ions in the solution causes the equilibrium to shift according to Le Chatelier's principle. In this case, the equilibrium shifts to the left (reactants), which means the solubility of \(\mathrm{PbI}_{2}\) decreases. As a result, more solid \(\mathrm{PbI}_{2}\) will precipitate, increasing the amount of solid \(\mathrm{PbI}_{2}\) at the bottom of the beaker. (a) The amount of solid \(\mathrm{PbI}_{2}\) at the bottom of the beaker will increase.
04

Analyzing the change in the \(\mathrm{Pb}^{2+}\) concentration

Since the equilibrium shifts to the left, the concentration of \(\mathrm{Pb}^{2+}\) will decrease as more \(\mathrm{PbI}_{2}\) precipitates. (b) The concentration of \(\mathrm{Pb}^{2+}\) ions in the solution will decrease.
05

Analyzing the change in the \(I^-\) concentration

As the concentration of \(I^-\) ions increases due to the dissolution of \(\mathrm{KI}\), the solubility of \(\mathrm{PbI}_{2}\) decreases. However, the overall concentration of the \(I^-\) ions in the solution will still increase because the dissociation of \(\mathrm{KI}\) contributes more \(I^-\) ions than the decrease in \(I^-\) ions due to the precipitation of \(\mathrm{PbI}_{2}\). (c) The concentration of \(I^-\) ions in the solution will increase.

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Most popular questions from this chapter

A solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added dropwise to a solution that is 0.010\(M\) in \(\mathrm{Ba}^{2+}(a q)\) and 0.010\(M\) in \(\mathrm{Sr}^{2+}(a q) .\) (a) What concentration of \(\mathrm{SO}_{4}^{2-}\) is necessary to begin precipitation? (Neglect volume changes. \(\mathrm{BaSO}_{4} : K_{s p}=1.1 \times 10^{-10} ; \mathrm{SrSO}_{4}\): \(K_{s p}=3.2 \times 10^{-7} . )\) (b) Which cation precipitates first? (c) What is the concentration of \(\mathrm{SO}_{4}^{2-}(a q)\) when the second cation begins to precipitate?

The solubility product for Zn (OH) \(_{2}\) is \(3.0 \times 10^{-16} .\) The formation constant for the hydroxo complex, Zn (OH) \(_{4}^{2-}\) , is 4.6 \(\times 10^{17} .\) What concentration of \(\mathrm{OH}^{-}\) is required to dissolve 0.015 mol of \(\mathrm{Zn}(\mathrm{OH})_{2}\) in a liter of solution?

A solution containing several metal ions is treated with dilute HCl; no precipitate forms. The \(\mathrm{pH}\) is adjusted to about \(1,\) and \(\mathrm{H}_{2} \mathrm{S}\) is bubbled through. Again, no precipitate forms. The \(\mathrm{pH}\) of the solution is then adjusted to about \(8 .\) Again, \(\mathrm{H}_{2} \mathrm{S}\) is bubbled through. This time a precipitate forms. The filtrate from this solution is treated with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4} .\) No precipitate forms. Which of these metal cations are either possibly present or definitely absent: \(\mathrm{Al}^{3+}, \mathrm{Na}^{+}, \mathrm{Ag}^{+}, \mathrm{Mg}^{2+} ?\)

Compare the titration of a strong, monoprotic acid with a strong base to the titration of a weak, monoprotic acid with a strong base. Assume the strong and weak acid solutions initially have the same concentrations. Indicate whether the following statements are true or false. (a) More base is required to reach the equivalence point for the strong acid than the weak acid. (b) The pH at the beginning of the titration is lower for the weak acid than the strong acid. (c) The pH at the equivalence point is 7 no matter which acid is titrated.

Which of the following salts will be substantially more soluble in acidic solution than in pure water: (a) ZnCO \(_{3}\) \(\mathbf{b} ) \mathrm{ZnS},(\mathbf{c}) \mathrm{BiI}_{3},(\mathbf{d}) \mathrm{AgCN},(\mathbf{e}) \mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2} ?\)
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