Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 87

On a hot day, a 200.0 -mL sample of a saturated solution of \(\mathrm{PbI}_{2}\) was allowed to evaporate until dry. If 240 mg of solid \(\mathrm{PbI}_{2}\) was collected after evaporation was complete, calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{PbI}_{2}\) on this hot day.

Short Answer

Expert verified
The Ksp value for PbI2 on this hot day is \(7.062\times10^{-9}\).

Step by step solution

01

Calculate the concentration of \(\mathrm{PbI}_{2}\) in the solution

First, we need to convert the mass of PbI2 collected (240 mg) into moles by using the molar mass of PbI2 (461 g/mol): $$\frac{240\,\text{mg}}{1}\times\frac{1\,\text{g}}{1000\,\text{mg}}\times\frac{1\,\text{mol}}{461\,\text{g}} = 5.206\times10^{-4}\,\text{mol}$$ Next, we determine the moles of PbI2 in the initial 200.0 mL solution by dividing the moles of PbI2 by the volume in liters: $$\frac{5.206\times10^{-4}\,\text{mol}}{0.2000\,\text{L}} = 2.603\times10^{-3}\,\text{M}$$ The concentration of PbI2 in the saturated solution is 2.603 x 10^-3 M.
02

Write the balanced equation and Ksp expression for \(\mathrm{PbI}_{2}\) dissolution

Write the balanced equation for the dissolution of PbI2 in water: $$\mathrm{PbI_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2\,I^-_{(aq)}}$$ Now, write the solubility product constant (Ksp) expression for the above reaction: $$K_\mathrm{sp} = [\mathrm{Pb^{2+}}][\mathrm{I^-}]^2$$
03

Determine the concentrations of Pb2+ and I- ions in the saturated solution

We can use stoichiometry to determine the concentrations of Pb2+ and I- ions in the saturated solution. For every mole of PbI2 dissolved, 1 mole of Pb2+ and 2 moles of I- ions are produced. Thus, the concentration of Pb2+ ions is equal to the concentration of PbI2 in the solution: $$[\mathrm{Pb^{2+}}] = 2.603\times10^{-3}\,\text{M}$$ The concentration of I- ions is twice the concentration of PbI2 in the solution: $$[\mathrm{I^-}] = 2\times(2.603\times10^{-3}\,\text{M}) = 5.206\times10^{-3}\,\text{M}$$
04

Calculate the Ksp value for \(\mathrm{PbI}_{2}\)

Now that we have the concentrations of Pb2+ and I- ions, we can use the Ksp expression to find the Ksp value for PbI2: $$K_\mathrm{sp} = [\mathrm{Pb^{2+}}][\mathrm{I^-}]^2 = (2.603\times10^{-3})(5.206\times10^{-3})^2$$ $$K_\mathrm{sp} = 7.062\times10^{-9}$$ The Ksp value for PbI2 on this hot day is 7.062 x 10^-9.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) in a \(0.10-M \mathrm{KIO}_{3}\) solution is $2.6 \times 10^{-11} \mathrm{mol} / \mathrm{L}\( . Calculate \)K_{\mathrm{sp}}$ for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}.\)

The solubility of \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) in a $0.20-M \mathrm{KIO}_{3}\( solution is \)4.4 \times 10^{-8} \mathrm{mol} / \mathrm{L}$ . Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}.\)

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.

Consider a solution made by mixing \(500.0 \mathrm{mL}\) of $4.0 \mathrm{M} \mathrm{NH}_{3}\( and \)500.0 \mathrm{mL}\( of \)0.40 \mathrm{M} \mathrm{AgNO}_{3} . \mathrm{Ag}^{+}\( reacts with \)\mathrm{NH}_{3}$ to form \(\mathrm{AgNH}_{3}^{+}\) and \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}:\) $$\mathrm{Ag}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{AgNH}_{3}^{+}(a q) \qquad K_{1}=2.1 \times 10^{3}$$ $$\operatorname{AgNH}_{3}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K_{2}=8.2 \times 10^{3}$$ Determine the concentration of all species in solution.

Calculate the equilibrium concentrations of $\mathrm{NH}_{3}, \mathrm{Cu}^{2+}\( \)\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+},$ and \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) in a solution prepared by mixing \(500.0 \mathrm{mL}\) of \(3.00M\) \(\mathrm{NH}_{3}\) with $500.0 \mathrm{mL}\( of \)2.00 \times 10^{-3} \mathrm{M} \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}$ . The step wise equilibria are $$\mathrm{Cu}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{CuNH}_{3}^{2+}(a q) \quad K_{1}=1.86 \times 10^{4}$$ $$\mathrm{CuNH}_{3}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}(a q) \quad K_{2}=3.88 \times 10^{3}$$ $$\mathrm{Cu}\left(\mathrm{NH}_{2}\right)_{2}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{2}\right)_{3}^{2+}(a q) \quad K_{3}=1.00 \times 10^{3}$$ $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K_{4}=1.55 \times 10^{2}$$
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free