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Chapter 16: Problem 18

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.

Short Answer

Expert verified
The dissolution of \(\mathrm{PbCl}_{2}(s)\) in water is endothermic. This is because the solubility of lead(II) chloride increases with an increase in temperature, which indicates that the process is absorbing heat.

Step by step solution

01

Recall the relationship between solubility, temperature, and heat of reaction

In a dissolution process, the solubility of a substance in a solvent is affected by temperature changes. The effect of temperature on solubility depends on the heat of reaction involved in the dissolution process. - If the dissolution process is exothermic (releases heat), increasing the temperature will decrease the solubility. - If the dissolution process is endothermic (absorbs heat), increasing the temperature will increase the solubility. This relationship is based on Le Chatelier's principle, which states that a system at equilibrium will shift to minimize the effects of a change in the system's conditions.
02

Analyze the given information

According to the problem, the solubility of lead(II) chloride (\(\mathrm{PbCl}_{2}\)) in water increases with an increase in temperature. This implies that the dissolution process for \(\mathrm{PbCl}_{2}\) must be absorbing heat.
03

Determine if the dissolution is exothermic or endothermic

Since the dissolution process of \(\mathrm{PbCl}_{2}\) in water is absorbing heat (temperature increase leads to an increase in solubility), the process is endothermic. Hence, the dissolution of \(\mathrm{PbCl}_{2}(s)\) in water is endothermic.

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Most popular questions from this chapter

The copper(l) ion forms a complex ion with \(\mathrm{CN}^{-}\) according to the following equation: $$\mathrm{Cu}^{+}(a q)+3 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Cu}(\mathrm{CN})_{3}^{2-}(a q) \quad K=1.0 \times 10^{11}$$ a. Calculate the solubility of $\mathrm{CuBr}(s)\left(K_{\mathrm{sp}}=1.0 \times 10^{-5}\right)\( in \)1.0 \mathrm{L}\( of \)1.0 \mathrm{M} \mathrm{NaCN}$ . b. Calculate the concentration of \(\mathrm{Br}^{-}\) at equilibrium. c. Calculate the concentration of \(\mathrm{CN}^{-}\) at equilibrium.

Magnesium hydroxide, \(\operatorname{Mg}(\mathrm{OH})_{2},\) is the active ingredient in the antacid TUMS and has a \(K_{\mathrm{sp}}\) value of $8.9 \times 10^{-12}\( . If a 10.0 -g sample of \)\mathrm{Mg}(\mathrm{OH})_{2}$ is placed in \(500.0 \mathrm{mL}\) of solution, calculate the moles of OH -ions present. Because the \(K_{\mathrm{sp}}\) value for \(\mathrm{Mg}(\mathrm{OH})_{2}\) is much less than \(1,\) not a lot solid dissolves in solution. Explain how \(\mathrm{Mg}(\mathrm{OH})_{2}\) works to neutralize large amounts of stomach acid.

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.

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. \(0.010 M\) \(\mathrm{NaI}\)

Two different compounds have about the same molar solubility. Do they also have about the same \(K_{\text {sp}}\) value?
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