Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 99

The solubility of \(\mathrm{CaCO}_{3}\) is pH dependent. (a) Calculate the molar solubility of \(\mathrm{CaCO}_{3}\left(K_{s p}=4.5 \times 10^{-9}\right)\) neglecting the acid-base character of the carbonate ion. (b) Use the \(K_{b}\) expression for the \(\mathrm{CO}_{3}^{2-}\) ion to determine the equilibrium constant for the reaction $$ \mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons $$ (c) If we assume that the only sources of $\mathrm{Ca}^{2+}, \mathrm{HCO}_{3}^{-},\( and \)\mathrm{OH}^{-}$ ions are from the dissolution of \(\mathrm{CaCO}_{3},\) what is the molar solubility of \(\mathrm{CaCO}_{3}\) using the equilibrium expression from part (b)? (d) What is the molar solubility of \(\mathrm{CaCO}_{3}\) at the \(\mathrm{pH}\) of the ocean (8.3)\(?(\mathbf{e})\) If the \(\mathrm{pH}\) is buffered at \(7.5,\) what is the molar solubility of \(\mathrm{CaCO}_{3} ?\)

Short Answer

Expert verified
The molar solubility of \(\mathrm{CaCO}_3\) neglecting the acid-base character of the carbonate ion is calculated using the given solubility product (\(K_{sp} = 4.5 \times 10^{-9}\)) and found to be \(x = 2.12 \times 10^{-5}\) M. When considering the acid-base character of the carbonate ion, the molar solubility of \(\mathrm{CaCO}_3\) is found using the Kb expression, and the specific pH values are considered. At a pH of 8.3 (ocean pH), the molar solubility of \(\mathrm{CaCO}_3\) is found to be higher than at a pH of 7.5 (buffered pH).

Step by step solution

01

Calculate the molar solubility neglecting the acid-base character of the carbonate ion

We are given the solubility product, \(K_{sp}\), for the dissolution of \(\mathrm{CaCO}_3\): $$\mathrm{CaCO_3} \leftrightarrow \mathrm{Ca^{2+}}+\mathrm{CO_3^{2-}}$$ and \(K_{sp} = 4.5\times10^{-9}\). Let the molar solubility of \(\mathrm{CaCO}_3\) be \(x\). Then, at equilibrium, the concentrations of \(\mathrm{Ca^{2+}}\) and \(\mathrm{CO_3^{2-}}\) would be \(x\) and \(x\) respectively. We can write the \(K_{sp}\) expression as $$K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{CO_3^{2-}}] = x^2$$ Solving for \(x\), we get the molar solubility.
02

Use the Kb expression for the CO3^2- ion to determine the equilibrium constant for the reaction

The Kb expression for the \(\mathrm{CO_3^{2-}}\) ion is given by: $$\mathrm{CO_3^{2-}}+\mathrm{H_2O} \rightleftharpoons \mathrm{HCO_3^-}+\mathrm{OH^-}$$ The equilibrium constant expression for the reaction of \(\mathrm{CaCO}_3\) with water can be represented as: $$K = \frac{[\mathrm{HCO_3^-}][\mathrm{OH^-}]}{[\mathrm{CO_3^{2-}}]}$$ We can replace \([\mathrm{HCO_3^-}]\) and \([\mathrm{OH^-}]\) with \(x\) since the molar solubility of \(\mathrm{CaCO}_3\) is assumed to be the only source of these ions. Since the concentration of the water is considered to be constant, we can include it in the equilibrium constant. Thus, the new equilibrium constant can be found.
03

Calculate the molar solubility of CaCO3 using the equilibrium expression from part (b)

Using the new equilibrium constant, we can write the equilibrium expression for the reaction involving \(\mathrm{CaCO}_3\) and water: $$K = \frac{x^2}{[\mathrm{CO_3^{2-}}]}$$ We can once again replace \([\mathrm{CO_3^{2-}}]\) with \(x\) since the molar solubility of \(\mathrm{CaCO}_3\) is assumed to be the only source of carbonate ions. Solving this equation for \(x\) will give us the molar solubility of \(\mathrm{CaCO}_3\) under these conditions.
04

Calculate the molar solubility of CaCO3 at the pH of the ocean (8.3)

To find the molar solubility of \(\mathrm{CaCO}_3\) at the pH of the ocean (8.3), we first need to calculate the concentration of \(\mathrm{CO_3^{2-}}\) at this pH. Using the relationship between pH, pOH, and pKw: $$\mathrm{pH}+\mathrm{pOH}=\mathrm{pK_w} \Rightarrow \mathrm{pOH}=\mathrm{pK_w}-\mathrm{pH}$$ We can find the concentration of \(\mathrm{OH^-}\) ions at pH 8.3 using pOH and then find the concentration of \(\mathrm{CO_3^{2-}}\) using the Kb expression for the \(\mathrm{CO_3^{2-}}\) ion. Finally, using the equilibrium expression derived in Step 3, we can find the molar solubility of \(\mathrm{CaCO}_3\) at this pH.
05

Calculate the molar solubility of CaCO3 at pH 7.5

Similar to Step 4, we need to find the concentration of \(\mathrm{CO_3^{2-}}\) ions at pH 7.5. Using the relationship between pH, pOH, and pKw, we can find the concentration of \(\mathrm{OH^-}\) ions at pH 7.5. Then, using the Kb expression for the \(\mathrm{CO_3^{2-}}\) ion, we can find the concentration of \(\mathrm{CO_3^{2-}}\). Finally, using the equilibrium expression derived in Step 3, we can find the molar solubility of \(\mathrm{CaCO}_3\) at pH 7.5.

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) A 0.1044-g sample of an unknown monoprotic acid requires $22.10 \mathrm{~mL}\( of \)0.0500 \mathrm{M} \mathrm{NaOH}$ to reach the end point. What is the molar mass of the unknown? (b) As the acid is titrated, the \(\mathrm{pH}\) of the solution after the addition of \(11.05 \mathrm{~mL}\) of the base is \(4.89 .\) What is the \(K_{a}\) for the acid? (c) Using Appendix D, suggest the identity of the acid.

How many milliliters of \(0.105 \mathrm{MHCl}\) are needed to titrate each of the following solutions to the equivalence point: (a) 45.0 \(\mathrm{mL}\) of \(0.0950 \mathrm{MNaOH}\) (b) \(22.5 \mathrm{~mL}\) of \(0.118 \mathrm{MNH}_{3}\), (c) 125.0 mL of a solution that contains \(1.35 \mathrm{~g}\) of \(\mathrm{NaOH}\) per liter?

Fluoridation of drinking water is employed in many places to aid in the prevention of tooth decay. Typically. the \(\mathrm{F}^{-}\) ion concentration is adjusted to about \(1 \mathrm{ppm}\). Some water supplies are also "hard"; that is, they contain certain cations such as \(\mathrm{Ca}^{2+}\) that interfere with the action of soap. Consider a case where the concentration of \(\mathrm{Ca}^{2+}\) is \(8 \mathrm{ppm}\). Could a precipitate of \(\mathrm{CaF}_{2}\) form under these conditions? (Make any necessary approximations.)

Suppose that a 10 -mL sample of a solution is to be tested for \(\mathrm{I}^{-}\) ion by addition of 1 drop \((0.2 \mathrm{~mL})\) of $0.10 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$ What is the minimum number of grams of \(\mathrm{I}^{-}\) that must be present for \(\mathrm{PbI}_{2}(s)\) to form?

Which of these statements about the common-ion effect is most correct? (a) The solubility of a salt MA is decreased in a solution that already contains either \(\mathrm{M}^{+}\) or \(\mathrm{A}^{-} .\) (b) Common ions alter the equilibrium constant for the reaction of an ionic solid with water. \((\mathbf{c})\) The common-ion effect does not apply to unusual ions like \(\mathrm{SO}_{3}^{2-} .(\mathbf{d})\) The solubility of a salt MA is affected equally by the addition of either \(\mathrm{A}^{-}\) or a noncommon ion.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

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