Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 4

Calculate the aqueous solubility, in moles per liter, of each of the following. (a) \(\mathrm{BaCrO}_{4}, K_{\mathrm{sp}}=1.2 \times 10^{-10}\) (b) \(\mathrm{PbBr}_{2}, K_{\mathrm{sp}}=4.0 \times 10^{-5}\) (c) \(\mathrm{CeF}_{3}, K_{\mathrm{sp}}=8 \times 10^{-16}\) (d) \(\operatorname{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}, K_{\mathrm{sp}}=2.1 \times 10^{-20}\)

Short Answer

Expert verified
Repeat these steps for each compound. The calculations yield the solubility as follows: (a) \(1.1 \times 10^{-5}\) M, (b) \(2.0 \times 10^{-2}\) M, (c) \(2.0 \times 10^{-5}\) M, (d) \(2.5 \times 10^{-7}\) M.

Step by step solution

01

Write the dissolution equation

Each compound will dissociate into ions in solution. For the first compound, \(\mathrm{BaCrO}_{4}\):\[\mathrm{BaCrO}_{4}(s) \leftrightarrow \mathrm{Ba}^{2+}(aq) + \mathrm{CrO}_4^{2-}(aq)\]
02

Establish the solubility product expression

The solubility product is determined by the multiplication of the concentrations of the ions derived from the compound, each raised to the power of its stoichiometric coefficient. For \(\mathrm{BaCrO}_{4}\), the expression is:\[[K_{sp}] = [\mathrm{Ba}^{2+}][\mathrm{CrO}_4^{2-}]\]
03

Determine the numerical value of the expression

Let \(x\) be the solubility of \(\mathrm{BaCrO}_{4}\) in moles per liter. Because of the 1:1 stoichiometric ratio, the concentrations of both ions in solution are equal to \(x\). Thus:\[[K_{sp}] = (x)(x) = \(x^2\) = 1.2 \times 10^{-10}\]
04

Solve for solubility

From the above equation, we can solve for \(x\), which is the solubility of \(\mathrm{BaCrO}_{4}\) in moles per liter, by taking the square root of both sides:\[x = \(\sqrt{1.2 \times 10^{-10}}\)\]

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

Will AgI(s) precipitate from a solution with \(\left[\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\right]=0.012 \mathrm{M}, \left[\mathrm{CN}^{-}\right]=1.05 \mathrm{M}, \) and \(\left[\mathrm{I}^{-}\right]=2.0 \mathrm{M} ?\) For \( \mathrm{AgI}, K_{\mathrm{sp}}=8.5 \times 10^{-17} ; =\) for \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}, K_{\mathrm{f}}=5.6 \times 10^{18}\).

A solution is \(0.010 \mathrm{M}\) in both \(\mathrm{CrO}_{4}^{2-}\) and \(\mathrm{SO}_{4}^{2-}\). To this solution, \(0.50 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\text { aq })\) is slowly added. (a) Which anion will precipitate first from solution? (b) What is \(\left[\mathrm{Pb}^{2+}\right]\) at the point at which the second anion begins to precipitate? (c) Are the two anions effectively separated by this fractional precipitation?

Excess \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})\) is shaken with water to produce a saturated solution. A 50.00 mL sample of the clear saturated solution is withdrawn and requires \(10.7 \mathrm{mL}\) of \(0.1032 \mathrm{M} \mathrm{HCl}\) for its titration. What is \(K_{\mathrm{sp}}\) for \(\mathrm{Ca}(\mathrm{OH})_{2} ?\)

Will the following precipitates form under the given conditions? (a) \(\mathrm{PbI}_{2}(\mathrm{s}),\) from a solution that is \(1.05 \times 10^{-3} \mathrm{M} \mathrm{HI}\), \(1.05 \times 10^{-3} \mathrm{M} \mathrm{NaI},\) and \(1.1 \times 10^{-3} \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\). (b) \(\operatorname{Mg}(\mathrm{OH})_{2}(\mathrm{s}),\) from \(2.50 \mathrm{L}\) of \(0.0150 \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) to which is added 1 drop \((0.05 \mathrm{mL})\) of \(6.00 \mathrm{M} \mathrm{NH}_{3}\). (c) \(\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})\) from a solution that is \(0.010 \mathrm{M}\) in \(\mathrm{Al}^{3+}, 0.010 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH},\) and \(0.010 \mathrm{M} \mathrm{NaCH}_{3} \mathrm{COO}\).

Write \(K_{\text {sp }}\) expressions for the following equilibria. For example, for the reaction \(\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}(\mathrm{aq})+\) \(\mathrm{Cl}^{-}(\mathrm{aq}), K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]\). (a) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (b) \(\operatorname{Ra}\left(\mathrm{IO}_{3}\right)_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Ra}^{2+}(\mathrm{aq})+2 \mathrm{IO}_{3}^{-}(\mathrm{aq})\) (c) \(\mathrm{Ni}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s}) \rightleftharpoons 3 \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{PO}_{4}^{3-}(\mathrm{aq})\) (d) \(\mathrm{PuO}_{2} \mathrm{CO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{PuO}_{2}^{2+}(\mathrm{aq})+\mathrm{CO}_{3}^{2-}(\mathrm{aq})\)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

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