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Chapter 16: Problem 44

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. \(0.10M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) c. \(0.010 M\) \(\mathrm{NaI}\)

Short Answer

Expert verified
The solubility of lead iodide (PbI2) in the given solutions is as follows: a. In water: approximately \(1.52 \times 10^{-3}\, \text{mol/L}\) b. In 0.10M Pb(NO3)2: approximately \(5.92 \times 10^{-4}\, \text{mol/L}\) c. In 0.010M NaI: approximately \(1.26 \times 10^{-6}\, \text{mol/L}\)

Step by step solution

01

Write the Dissociation Reaction

First, we need to write a balanced dissociation reaction for lead iodide (PbI2). This will allow us to find the relationship between the concentrations of lead and iodide ions in the solution. The dissociation reaction is: \[ \mathrm{PbI}_{2} (s) \rightleftharpoons \mathrm{Pb^{2+}} (aq) + 2 \mathrm{I^{-}} (aq) \]
02

Write the Ksp Expression

Now, we can write the Ksp expression for the given reaction. For the dissociation reaction, Ksp is given by the product of the concentrations of the ions involved, raised to the power of their stoichiometric coefficients. Ksp expression is: \[ K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{I^{-}}]^2 \]
03

Calculate Solubility in Water (a)

Let the solubility of PbI2 in water be 's' mol/L. According to the dissociation reaction, when 1 mol of PbI2 dissolves, 1 mol of Pb2+ and 2 mol of I- are formed. Therefore, the concentration of Pb2+ will also be s, while the concentration of I- will be 2s. Substitute the concentrations in the Ksp expression and solve for s: \[ K_{sp} = (s)(2s)^2 \] \[ 1.4 \times 10^{-8} = (s)(4s^2)\] \[ s^3 = 3.5 \times 10^{-9} \] \[ s = \sqrt[3]{3.5 \times 10^{-9}} \approx 1.52 \times 10^{-3}\,\text{mol/L}\] The solubility of lead iodide in water is approximately \(1.52 \times 10^{-3}\, \text{mol/L}\).
04

Calculate Solubility in 0.10M Pb(NO3)2 (b)

We are given [Pb2+] as 0.10M, and let the increase in [I-] due to the solubility be s. Since we already have [Pb2+], the expression for Ksp can be simplified. \[ K_{sp} = (0.10)(2s)^2 \] \[ 1.4 \times 10^{-8} = (0.10)(4s^2)\] \[ s^2 = 3.5 \times 10^{-8} \] \[ s = \sqrt{3.5 \times 10^{-8}} \approx 5.92 \times 10^{-4}\,\text{mol/L}\] The solubility of lead iodide in 0.10M Pb(NO3)2 solution is approximately \(5.92 \times 10^{-4}\, \text{mol/L}\)
05

Calculate Solubility in 0.010M NaI (c)

We are given [I-] as 0.010M, and let the increase in [Pb2+] due to the solubility be s. Since we already have [I-], the expression for Ksp can be simplified. \[ K_{sp} = (s)(0.010 + 2s)^2 \] \[ 1.4 \times 10^{-8} = s(0.010^2 + 0.04s + 4s^2) \] Simplifying the equation, we can neglect the term s^2: \[ 1.4 \times 10^{-8} \approx s(0.010^2 + 0.04s)\] Solve for s: \[ s \approx \frac{1.4 \times 10^{-8}}{0.010^2 + 0.04s} \approx 1.26 \times 10^{-6}\, \text{mol/L} \] The solubility of lead iodide in 0.010M NaI solution is approximately \(1.26 \times 10^{-6}\, \text{mol/L}\).

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Most popular questions from this chapter

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

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