Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 8

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

Short Answer

Expert verified
Two different compounds having about the same molar solubility do not necessarily imply that they have the same Ksp values. However, if their molar solubility is similar (s1 ≈ s2), then their Ksp values might also be similar (Ksp(AB) ≈ Ksp(CD)) as Ksp(AB) = \(s_1^2\) and Ksp(CD) = \(s_2^2\). This is just a general observation and might not hold true for every pair of compounds with similar molar solubility. It is important to evaluate each specific case for a more accurate comparison.

Step by step solution

01

Define molar solubility and solubility product constant (Ksp)

Molar solubility is the number of moles of a solute that can dissolve in one liter of a solution to form a saturated solution. The solubility product constant (Ksp) is the equilibrium constant for a chemical reaction in which a solid ionic compound dissolves to yield its ions in a solution. It provides a quantitative measure of the solubility of the compound.
02

Express the solubility equilibrium and Ksp expressions for two generic compounds

Let's represent the two compounds as AB (with the molar solubility s1) and CD (with the molar solubility s2). The solubility equilibrium for these compounds can be represented as: Compound AB: \(AB_{(s)} \rightleftharpoons A_{(aq)}^+ + B_{(aq)}^-\) Compound CD: \(CD_{(s)} \rightleftharpoons C_{(aq)}^{+} + D_{(aq)}^{-}\) Now we can write the Ksp expressions for these compounds: Ksp(AB) = [A+][B−] = \(s_1^2\) Ksp(CD) = [C+][D−] = \(s_2^2\)
03

Analyze the relationship between molar solubility and Ksp

Assuming that the two compounds have similar molar solubility (s1 ≈ s2), we can examine the relationship between their Ksp values. Since Ksp(AB) = \(s_1^2\) and Ksp(CD) = \(s_2^2\), if s1 ≈ s2, then it implies that Ksp(AB) ≈ Ksp(CD). Therefore, if two different compounds have about the same molar solubility, they are likely to have about the same Ksp values. However, this is just a general observation and might not hold true for every pair of compounds with similar molar solubility. It is important to evaluate each specific case for a more accurate comparison.

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

a. Calculate the molar solubility of \(\mathrm{AgBr}\) in pure water. \(K_{\mathrm{sp}}\) for \(\mathrm{AgBr}\) is \(5.0 \times 10^{-13}\) . b. Calculate the molar solubility of \(\mathrm{AgBr}\) in \(3.0M\) \(\mathrm{NH}_{3} .\) The overall formation constant for \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+\) is \(1.7 \times 10^{7}\) that is, $$\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}$$ c. Compare the calculated solubilities from parts a and b. Explain any differences. d. What mass of \(\mathrm{AgBr}\) will dissolve in \(250.0 \mathrm{mL}\) of $3.0 M\( \)\mathrm{NH}_{3}?$ e. What effect does adding \(\mathrm{HNO}_{3}\) have on the solubilities calculated in parts a and b?

What mass of \(\mathrm{ZnS}\left(K_{\mathrm{sp}}=2.5 \times 10^{-22}\right)\) will dissolve in 300.0 \(\mathrm{mL}\) of \(0.050M\) \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2} ?\) Ignore the basic properties of \(\mathrm{S}^{2-}.\)

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 \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}.\)

Sulfide precipitates are generally grouped as sulfides insoluble in acidic solution and sulfides insoluble in basic solution. Explain why there is a difference between the two groups of sulfide precipitates.

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.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Kinetics

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