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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 KI 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 \(PbI_{2}\) at the bottom of the beaker will increase, because the equilibrium shifts to the left due to the increased I⁻ ion concentration from the added KI. (b) The concentration of \(Pb^{2+}\) ions in the solution will decrease as the equilibrium shifts to the left. (c) The concentration of I⁻ ions in the solution will initially increase due to the addition of KI but may decrease as more solid \(PbI_{2}\) forms; however, the final concentration might still be higher than the initial one.

Step by step solution

01

Write the equation for the system

The system is given by the equilibrium between solid PbI₂ and its ions in water: \[PbI_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2 I^{-}(aq)\]
02

Write the equation for the addition of KI

Adding solid KI to the solution implies the dissolution of KI in water: \[KI(s) \rightarrow K^{+}(aq) + I^{-}(aq)\]
03

Apply the principle of Le Châtelier

The principle of Le Châtelier states that when a system in equilibrium is subjected to a change, the system will adjust the positions of the equilibrium to counteract the change. In this case, the addition of KI increases the concentration of I⁻ ions.
04

Analyze the effect on the amount of solid PbI2

As the concentration of I⁻ ions increases, the equilibrium for the PbI₂ system will shift to the left to counteract the change, according to Le Châtelier's principle. This means more solid PbI₂ will be formed, so (a) the amount of solid PbI₂ at the bottom of the beaker will increase.
05

Analyze the effect on the concentration of Pb²⁺ ions

Since the system is shifting to the left, due to the increased I⁻ ion concentration from the added KI, the concentration of Pb²⁺ ions will decrease to maintain the equilibrium. So, (b) the concentration of Pb²⁺ ions in the solution will decrease.
06

Analyze the effect on the concentration of I⁻ ions

Initially, the concentration of I⁻ ions will increase due to the addition of KI. However, as the system tries to counteract this change, the equilibrium shifts to the left, promoting the formation of solid PbI₂ by consuming I⁻ ions. Therefore, (c) the concentration of I⁻ ions in the solution will increase initially, and then likely decrease as more solid PbI₂ forms but may still be higher than the initial concentration.

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Most popular questions from this chapter

Suggest how the cations in each of the following solution mixtures can be separated: (a) \(\mathrm{Na}^{+}\) and $\mathrm{Cd}^{2+},(\mathbf{b}) \mathrm{Cu}^{2+}\( and \)\mathrm{Mg}^{2+},(\mathbf{c}) \mathrm{Pb}^{2+}$ and \(\mathrm{Al}^{3+},(\mathbf{d}) \mathrm{Ag}^{+}\) and \(\mathrm{Hg}^{2+} .\)

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A concentration of \(10-100\) parts per billion (by mass) of \(\mathrm{Ag}^{+}\) is an effective disinfectant in swimming pools. However, if the concentration exceeds this range, the \(\mathrm{Ag}^{+}\) can cause adverse health effects. One way to maintain an appropriate concentration of \(\mathrm{Ag}^{+}\) is to add a slightly soluble salt to the pool. Using \(K_{s p}\) values from Appendix \(D\), calculate the equilibrium concentration of \(\mathrm{Ag}^{+}\) in parts per billion that would exist in equilibrium with (a) AgCl, (b) AgBr, (c) AgI.

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