Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 45

Which of the following solids is (are) more soluble in an acidic solution than in pure water: \(\mathrm{KCl}\), \(\mathrm{MgCO}_{3}\), \(\mathrm{FeS}, \mathrm{Ca}(\mathrm{OH})_{2,}\) or \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} ?\) Explain.

Short Answer

Expert verified
\( \mathrm{MgCO}_3 \), \( \mathrm{FeS} \), and \( \mathrm{Ca(OH)}_2 \) are more soluble in acidic solution than in pure water because these have the ability to react with the acid to increase its solubility.

Step by step solution

01

Identifying the nature of each molecule

Let's start by categorizing these molecules: \n \( \mathrm{KCl} \) is a neutral salt and won't be affected by the acidity of the solution. \n \( \mathrm{MgCO}_3 \) is a basic salt, anions of which will react with the H+ ions present in the acid. \n \( \mathrm{FeS} \) is an amphoteric compound, can show both acidic and basic properties. It reacts with acid and forming H2S. \n \( \mathrm{Ca(OH)}_2 \) is a strong base and it will react with extra H+ ions present in the acid. \n \( \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COOH} \) is an acid and it won't react with another acid to raise its solubility.
02

Categorizing the compounds

Now based on step 1, we can categorize the molecules which are more soluble in acid than in water are \( \mathrm{MgCO}_3 \), \( \mathrm{FeS} \), and \( \mathrm{Ca(OH)}_2 \) because these are the only compounds among these that can react with acids to increase its solubility.

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 electrolysis of \(\mathrm{MgCl}_{2}(\mathrm{aq})\) can be represented as \(\mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1) \longrightarrow\) \(\mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) The electrolysis of a 315 mL sample of \(0.185 \mathrm{M} \mathrm{MgCl}_{2}\) is continued until \(0.652 \mathrm{L} \mathrm{H}_{2}(\mathrm{g})\) at \(22^{\circ} \mathrm{C}\) and \(752 \mathrm{mmHg}\) has been collected. Will \(\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})\) precipitate when electrolysis is carried to this point? [Hint: Notice that \(\left[\mathrm{Mg}^{2+}\right]\) remains constant throughout the electrolysis, but \(\left.\left[\mathrm{OH}^{-}\right] \text {increases. }\right]\)

\(\mathrm{KI}(\mathrm{aq})\) is slowly added to a solution with \(\left[\mathrm{Pb}^{2+}\right]=\) \(\left[\mathrm{Ag}^{+}\right]=0.10 \mathrm{M} .\) For \(\mathrm{PbI}_{2}, K_{\mathrm{sp}}=7.1 \times 10^{-9} ;\) for \(\mathrm{AgI},\) \(K_{\mathrm{sp}}=8.5 \times 10^{-17}\). (a) Which precipitate should form first, \(\mathrm{PbI}_{2}\) or AgI? (b) What \(\left[\mathrm{I}^{-}\right]\) is required for the second cation to begin to precipitate? (c) What concentration of the first cation to precipitate remains in solution at the point at which the second cation begins to precipitate? (d) \(\operatorname{Can} \mathrm{Pb}^{2+}(\mathrm{aq})\) and \(\mathrm{Ag}^{+}(\) aq) be effectively separated by fractional precipitation of their iodides?

Write net ionic equations for each of the following observations. (a) When concentrated \(\mathrm{CaCl}_{2}(\mathrm{aq})\) is added to \(\mathrm{Na}_{2} \mathrm{HPO}_{4}(\mathrm{aq}),\) a white precipitate forms that is \(38.7 \%\) Ca by mass. (b) When a piece of dry ice, \(\mathrm{CO}_{2}(\mathrm{s}),\) is placed in a clear dilute solution of limewater \(\left[\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq})\right]\), bubbles of gas evolve. At first, a white precipitate forms, but then it redissolves.

Calculate \(\left[\mathrm{Cu}^{2+}\right]\) in a \(0.10 \mathrm{M} \mathrm{CuSO}_{4}(\) aq) solution that is also \(6.0 \mathrm{M}\) in free \(\mathrm{NH}_{3}\). \(\mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}(\mathrm{aq})\) \(K_{\mathrm{f}}=1.1 \times 10^{13}\)

Both \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cu}^{2+}\) are present in the same aqueous solution. Which of the following reagents would work best in separating these ions, precipitating one and leaving the other in solution: \(\mathrm{NaOH}(\mathrm{aq}), \mathrm{HCl}(\mathrm{aq})\), \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq}),\) or \(\mathrm{NH}_{3}(\mathrm{aq}) ?\) Explain your choice.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

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