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Chapter 16: Problem 29

The concentration of \(\mathrm{Pb}^{2+}\) in a solution saturated with \(\mathrm{PbBr}_{2}(s)\) is \(2.14 \times 10^{-2} \mathrm{M} .\) Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{PbBr}_{2}.\)

Short Answer

Expert verified
The solubility product constant (\(K_\text{sp}\)) for PbBr2 is \(1.96 \times 10^{-5}\).

Step by step solution

01

Write the balanced chemical equation for the dissociation of PbBr2 in water.

PbBr2 (s) ⇌ Pb2+ (aq) + 2 Br- (aq)
02

Write the expression for the solubility product constant, Ksp.

According to the balanced chemical equation, the Ksp expression for the reaction can be written as follows: \[K_\text{sp} = [Pb^{2+}][Br^-]^2\]
03

Determine the equilibrium concentrations of the ions.

The problem states that the concentration of Pb2+ in a saturated solution is \(2.14 \times 10^{-2} M\). Since the balanced chemical equation shows a 1:2 ratio between Pb2+ and Br-, the equilibrium concentration of Br- ions must be twice that of Pb2+ ions. Therefore, the concentration of Br- is \(2(2.14 \times 10^{-2} M) = 4.28 \times 10^{-2} M\).
04

Calculate Ksp using the equilibrium concentrations.

Plug the equilibrium concentrations of Pb2+ and Br- into the Ksp expression: \[K_\text{sp} = [(2.14 \times 10^{-2})][(4.28 \times 10^{-2})^2]\] Now, to find the value of Ksp, simply evaluate this expression: \[K_\text{sp} = (2.14 \times 10^{-2}) \times (4.28 \times 10^{-2})^2\] \[K_\text{sp} = 1.96 \times 10^{-5}\] Therefore, the solubility product constant (\(K_\text{sp}\)) for PbBr2 is \(1.96 \times 10^{-5}\).

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Most popular questions from this chapter

The solubility of \(\mathrm{PbCl}_{2}\) increases with an increase in temperature. Is the dissolution of \(\mathrm{PbCl}_{2}(s)\) in water exothermic or endothermic? Explain.

A friend tells you: "The constant \(K_{\mathrm{sp}}\) of a salt is called the solubility product constant and is calculated from the concentrations of ions in the solution. Thus, if salt A dissolves to a greater extent than salt \(\mathrm{B}\) , salt \(\mathrm{A}\) must have a higher \(K_{\mathrm{sp}}\) than salt \(\mathrm{B}\) ." Do you agree with your friend? Explain.

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. \(\mathrm{BaF}_{2}\)

The salt MX has a solubility of $3.17 \times 10^{-8} \mathrm{mol} / \mathrm{L}\( in a solution with \)\mathrm{pH}=0.000 .\( If \)K_{\mathrm{a}}$ for \(\mathrm{HX}\) is \(1.00 \times 10^{-15}\) , calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{MX}\) .

The \(K_{\mathrm{sp}}\) for silver sulfate $\left(\mathrm{Ag}_{2} \mathrm{SO}_{4}\right)\( is \)1.2 \times 10^{-5} .$ Calculate the solubility of silver sulfate in each of the following. a. water b. \(0.10M\) \(\mathrm{AgNO}_{3}\) c. \(0.20M\) \(\mathrm{K}_{2} \mathrm{SO}_{4}\)
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