Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 58

At \(25^{\circ} \mathrm{C}\) the reaction $$\mathrm{CaCrO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CrO}_{4}^{2-}(a q)$$ has an equilibrium constant \(K_{c}=7.1 \times 10^{-4}\). What are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{CrO}_{4}^{2-}\) in a saturated solution of \(\mathrm{CaCrO}_{4} ?\)

Short Answer

Expert verified
The equilibrium concentrations of both \(Ca^{2+}\) and \(CrO_4^{2-}\) ions in the saturated solution of CaCrO₄ are 0.0266 M.

Step by step solution

01

Write the equilibrium expression

First, let's write the equilibrium expression for the given reaction. The equilibrium constant, Kc, is the product of the concentrations of the products divided by the product of the concentrations of the reactants. Based on the balanced chemical equation, the equilibrium expression is: \[K_c = \frac{[\mathrm{Ca}^{2+}][\mathrm{CrO}_{4}^{2-}]}{[\mathrm{CaCrO}_4]}\] Since CaCrO₄ is a solid, its concentration remains unchanged, and we can rewrite the equilibrium expression as: \[K_c = [\mathrm{Ca}^{2+}][\mathrm{CrO}_{4}^{2-}]\]
02

Set up the ICE table

Now, let's set up an ICE (Initial, Change, Equilibrium) table to keep track of the changes in the concentration of the species involved in the reaction. ``` CaCrO₄(s) ⟷ Ca²⁺(aq) + CrO₄²⁻(aq) Initial: - 0 0 Change: - +x +x Equilibrium: - x x ``` In the table, x represents the change in the concentration of each aqueous ion. Note that the concentration of the solid CaCrO₄ remains unchanged.
03

Substitute equilibrium concentrations into the equilibrium expression and solve for x

Now we can substitute the equilibrium concentrations from the ICE table into the equilibrium expression: \[K_c = [\mathrm{Ca}^{2+}][\mathrm{CrO}_{4}^{2-}] = (x)(x) = x^2\] Given the value of Kc: \[7.1 \times 10^{-4} = x^2\] Next, we will find the value of x by taking the square root of both sides: \[x = \sqrt{7.1 \times 10^{-4}}\] Calculating x, we get: \[x = 0.0266\]
04

Determine the equilibrium concentrations of Ca²⁺ and CrO₄²⁻

Now that we have the value of x, we can determine the equilibrium concentrations of Ca²⁺ and CrO₄²⁻. From the ICE table, we know that: \[[\mathrm{Ca}^{2+}] = x = 0.0266\] \[[\mathrm{CrO}_{4}^{2-}] = x = 0.0266\] Therefore, the equilibrium concentrations of Ca²⁺ and CrO₄²⁻ are both 0.0266 M.

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

If \(K_{c}=1\) for the equilibrium \(2 \mathrm{~A}(g) \rightleftharpoons \mathrm{B}(g)\), what is the relationship between \([\mathrm{A}]\) and \([\mathrm{B}]\) at equilibrium?

Explain why we normally exclude pure solids and liquids from equilibrium- constant expressions.

A mixture of \(\mathrm{H}_{2}, \mathrm{~S},\) and \(\mathrm{H}_{2} \mathrm{~S}\) is held in a \(1.0-\mathrm{L}\) vessel at \(90{ }^{\circ} \mathrm{C}\) and reacts according to the equation: $$\mathrm{H}_{2}(g)+\mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)$$ At equilibrium the mixture contains \(0.46 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and \(0.40 \mathrm{~g}\) \(\mathrm{H}_{2}\) (a) Write the equilibrium-constant expression for this reaction. (b) What is the value of \(K_{c}\) for the reaction at this temperature? (c) Why can we ignore the amount of \(\mathrm{S}\) when doing the calculation in part (b)?

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\). A 7.5-L gas vessel is charged with a mixture of \(\mathrm{PCl}_{3}(g)\) and \(\mathrm{Cl}_{2}(g),\) which is allowed to equilibrate at \(450 \mathrm{~K}\). At equilibrium the partial pressures of the three gases are \(P_{\mathrm{PCl}_{3}}=0.124 \mathrm{~atm}\) \(P_{\mathrm{Cl}_{2}}=0.157 \mathrm{~atm},\) and \(P_{\mathrm{PCl}_{5}}=1.30 \mathrm{~atm}\) (a) What is the value of \(K_{p}\) at this temperature? (b) Does the equilibrium favor reactants or products? (c) Calculate \(K_{c}\) for this reaction at \(450 \mathrm{~K}\).

If \(K_{c}=0.042\) for \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) at \(500 \mathrm{~K}\) what is the value of \(K_{p}\) for this reaction at this temperature?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

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