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Chapter 15: Problem 40

The \(K_{\mathrm{sp}}\) for lead iodide \(\left(\mathrm{PbI}_{2}\right)\) is \(1.4 \times 10^{-8} .\) Calculate the solubility of lead iodide in each of the following. a. water b. \(0.10 M \operatorname{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) c. \(0.010 M\) NaI

Short Answer

Expert verified
The solubility of PbI鈧� in the different solutions is as follows: a. In water, the molar solubility is approximately \(6.90 \times 10^{-3} M\) b. In \(0.10 M\) Pb(NO鈧�)鈧�, the molar solubility is approximately \(1.2 \times 10^{-7} M\) c. In \(0.010 M\) NaI, the molar solubility is approximately \(1.32 \times 10^{-4} M\) The presence of other ions in the solution affects the solubility of PbI鈧�, with solubility decreasing in the presence of Pb虏鈦� ions and increasing in the presence of I鈦� ions.

Step by step solution

01

Calculate the solubility of PbI鈧� in water

Dissolving PbI鈧� in water can be represented by the following equilibrium: \[PbI_2_{(s)} \rightleftharpoons \mathrm{Pb^{2+}}_{(aq)} + 2 \mathrm{I-}_{(aq)}\] To solve for the molar solubility of PbI鈧�, let x represent the solubility of PbI鈧� in molar units. Then, \[[Pb^{2+}]=[x] \\ [I^{-}]=2[x].\] Now, we can write the Ksp expression for this equilibrium as follows: \[K_{sp} = [Pb^{2+}][I^{-}]^{2}.\] Plugging in the values of the concentrations calculated above, we get \[1.4 \times 10^{-8} = [x](2x)^2.\] Now, solve for x, which represents the molar solubility of PbI鈧� in water.
02

Calculate the solubility of PbI鈧� in 0.10 M Pb(NO鈧�)鈧�

When PbI鈧� is dissolved in \(0.10 M\) Pb(NO鈧�)鈧�, the initial concentration of Pb虏鈦� ions is \(0.10 M\). Let y represent the solubility of PbI鈧� in the presence of Pb(NO鈧�)鈧�. The concentration of the ions will be: \[[Pb^{2+}] = 0.10 + y \\ [I^{-}] = 2y\] The Ksp expression for this equilibrium remains the same: \[K_{sp} = [Pb^{2+}][I^{-}]^{2}.\] Now, substitute the values of the concentrations into this expression: \[1.4 \times 10^{-8} = (0.10 + y)(2y)^2.\] Solve for y, which represents the molar solubility of PbI鈧� in \(0.10 M\) Pb(NO鈧�)鈧�.
03

Calculate the solubility of PbI鈧� in 0.010 M NaI

When PbI鈧� is dissolved in \(0.010 M\) NaI, the initial concentration of I鈦� ions is \(0.010 M\). Let z represent the solubility of PbI鈧� in the presence of NaI. The concentration of the ions will be: \[[Pb^{2+}] = z \\ [I^{-}] = 0.010 + 2z\] The Ksp expression for this equilibrium still applies: \[K_{sp} = [Pb^{2+}][I^{-}]^{2}.\] Now, substitute the values of the concentrations into this expression: \[1.4 \times 10^{-8} = z(0.010 + 2z)^2.\] Solve for z, representing the molar solubility of PbI鈧� in \(0.010 M\) NaI. For each part of the exercise, you'll find the molar solubility of PbI鈧� in the different solutions, which helps understand how the presence of other ions in the solution can affect the solubility of a compound.

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Key Concepts

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

Solubility Product Constant (Ksp)
The solubility product constant, commonly referred to with the symbol Ksp, is a special kind of equilibrium constant that applies to the solubility of ionic compounds. It represents the level at which a solute's ions are saturated in a solution, thus not allowing more solute to dissolve.

For lead iodide \(\mathrm{PbI}_{2}\), the equilibrium of dissolution can be written as: \[\mathrm{PbI}_{2\mathrm{(s)}} \rightleftharpoons \mathrm{Pb}^{2+}_{\mathrm{(aq)}} + 2 \mathrm{I}^-_{\mathrm{(aq)}}.\] The concentrations of the dissolved ions can be related to the Ksp by the expression \[K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{I}^-]^2.\] Here, the square brackets indicate the molar concentration of the ions at equilibrium. By setting up this relationship, we can determine the molar solubility of the compound in pure water or any aqueous solution.

In a step-by-step process, one first establishes this equilibrium expression, then identifies the solubility as 'x' for the lead cation and '2x' for the iodide anion, as the stoichiometry dictates there will be twice as much iodide. By substituting into the Ksp equation and solving for 'x,' we find the molar solubility of the compound in water.
Common Ion Effect
The common ion effect occurs when a compound is dissolved in a solution that already contains one of its constituent ions. This phenomenon affects the solubility of the compound, typically resulting in a decrease because the presence of the common ion shifts the equilibrium to the left, as per Le Chatelier's principle.

In the given example of lead iodide, when dissolving \(\mathrm{PbI}_{2}\) in a solution containing \(\mathrm{Pb}^{2+}\) ions from \(\mathrm{Pb(NO}_{3})_{2}\), the initial concentration of lead affects the solubility. Represented by 'y,' the new molar solubility must consider the initial concentration of lead ions, which modifies the equilibrium expression to \[K_{sp} = (0.10 + y)(2y)^2.\] Solving for 'y' gives the new solubility in the presence of a common ion.

Similarly, when \(\mathrm{PbI}_{2}\) is placed in a solution with sodium iodide (NaI), the common ion here is iodide \(\mathrm{I}^-\). The presence of this common ion will also decrease the solubility of \(\mathrm{PbI}_{2}\) as it shifts the dissolution equilibrium.
Equilibrium Concentration Calculations
To calculate equilibrium concentrations, it's important to begin with the balanced chemical equation which provides the stoichiometric ratios of the reactants and products. Then, we apply the solubility product constant equation to these concentrations.

In the context of the lead iodide case, we start with a given Ksp value and establish a system of equations based on the different scenarios: in water, with \(\mathrm{Pb(NO}_{3})_{2}\), or with NaI. The equilibrium concentration calculations then involve algebraic manipulation to solve for the unknown solubilities, designated as 'x,' 'y,' or 'z' for each respective situation.

The calculations take into account initial concentrations and any shifts in equilibrium due to the common ion effect. By applying these principles, students can understand how various factors, including the initial concentrations of other ions in solution, impact the solubility of ionic compounds.

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Most popular questions from this chapter

The equilibrium constant for the following reaction is \(1.0 \times 10^{23}:\) $$\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q) \rightleftharpoons \mathrm{CrEDTA}^{-}(a q)+2 \mathrm{H}^{+}(a q)$$ EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt \(\mathrm{Na}_{2} \mathrm{H}_{2} \mathrm{EDTA},\) are used to treat heavy metal poisoning. Calculate \(\left[\mathrm{Cr}^{3+}\right]\) at equilibrium in a solution originally \(0.0010 \mathrm{M}\) in \(\mathrm{Cr}^{3+}\) and \(0.050 M\) in \(\mathrm{H}_{2} \mathrm{EDTA}^{2-}\) and buffered at \(\mathrm{pH}=6.00\).

Write balanced equations for the dissolution reactions and the corresponding solubility product expressions for each of the following solids. a. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) b. \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) c. \(\mathrm{BaF}_{2}\)

When \(\mathrm{Na}_{3} \mathrm{PO}_{4}(a q)\) is added to a solution containing a metal ion and a precipitate forms, the precipitate generally could be one of two possibilities. What are the two possibilities?

A solution contains 0.018 molel each of \(\mathrm{I}^{-}, \mathrm{Br}^{-},\) and \(\mathrm{Cl}^{-}\). When the solution is mixed with \(200 . \mathrm{mL}\) of \(0.24\) \(M\) \(\mathrm{AgNO}_{3}\), what mass of \(\mathrm{AgCl}(s)\) precipitates out, and what is \(\left[\mathrm{Ag}^{+}\right] ?\) Assume no volume change. $$\begin{aligned} \operatorname{AgI}: K_{\mathrm{sp}} &=1.5 \times 10^{-16} \\ \operatorname{AgBr}: K_{\mathrm{sp}} &=5.0 \times 10^{-13} \\ \mathrm{AgCl}: K_{\mathrm{sp}} &=1.6 \times 10^{-10} \end{aligned}$$

a. Calculate the molar solubility of AgBr in pure water. \(K_{\mathrm{sp}}\) for AgBr is \(5.0 \times 10^{-13}\). b. Calculate the molar solubility of \(\mathrm{AgBr}\) in \(3.0\) \(M\) \(\mathrm{NH}_{3} .\) The overall formation constant for \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) is \(1.7 \times 10^{7}\) that is, $$\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}$$ c. Compare the calculated solubilities from parts a and b. Explain any differences. d. What mass of AgBr will dissolve in \(250.0 \mathrm{mL}\) of \(3.0 M \mathrm{NH}_{3} ?\) e. What effect does adding \(\mathrm{HNO}_{3}\) have on the solubilities calculated in parts a and b?
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