Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 85

$$K=\frac{\left[\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}\right]}{\left[\mathrm{Mn}^{2+}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]^{2}}$$

Short Answer

Expert verified
The equilibrium constant expression for the reaction between manganese(II) ions and oxalate ions forming manganese(II) tartrate ion is given by: \[K=\frac{\left[\mathrm{Mn}\left(\mathrm{C}_{2}\mathrm{O}_{4}\right)_{2}^{2-}\right]}{\left[\mathrm{Mn}^{2+}\right]\left[\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\right]^{2}}\] The value of K helps us understand the position of the chemical equilibrium, indicating whether the reaction favors the formation of manganese(II) tartrate ion, manganese(II) ions, and oxalate ions, or if the concentrations of reactants and products are equal at equilibrium.

Step by step solution

01

Understand the equilibrium constant expression

The equilibrium constant, K, is used to describe the relationship between the concentrations of reactants and products at equilibrium for a chemical reaction. It is an important parameter that helps us evaluate how "far" a reaction proceeds or how "favorable" the reaction is based on the concentrations of the species involved. In this expression, K is defined as the ratio of the concentration of manganese(II) tartrate ion to the concentrations of manganese(II) ions and oxalate ions squared.
02

Identify the chemical species involved in the expression

We have three chemical species involved in the expression: 1) Manganese(II) tartrate ion, \(\mathrm{Mn}\left(\mathrm{C}_{2}\mathrm{O}_{4}\right)_{2}^{2-}\), 2) Manganese(II) ions, \(\mathrm{Mn}^{2+}\), and 3) Oxalate ions, \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\).
03

Write the balanced chemical equation for the reaction

Based on the K expression, it can be concluded that this is a reaction involving one manganese(II) ion and two oxalate ions forming one manganese(II) tartrate ion. The balanced chemical equation can be written as follows: \[\mathrm{Mn}^{2+} + 2\mathrm{C}_{2}\mathrm{O}_{4}^{2-} \rightleftharpoons \mathrm{Mn}\left(\mathrm{C}_{2}\mathrm{O}_{4}\right)_{2}^{2-}\]
04

Apply the Law of Mass Action to the balanced chemical equation

According to the law of mass action, the equilibrium constant expression K is defined as the product of the concentrations of the products, raised to their stoichiometric coefficients, divided by the product of the concentrations of the reactants, raised to their stoichiometric coefficients. For our balanced chemical equation, the expression for K can be written as: \[K=\frac{\left[\mathrm{Mn}\left(\mathrm{C}_{2}\mathrm{O}_{4}\right)_{2}^{2-}\right]}{\left[\mathrm{Mn}^{2+}\right]\left[\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\right]^{2}}\]
05

Interpret the meaning of K

The value of K gives us information about the position of the chemical equilibrium. If K is greater than 1, the reaction favors the formation of the manganese(II) tartrate ion. If K is less than 1, the reaction favors the formation of manganese(II) ions and oxalate ions. A K value equal to 1 indicates that the reaction is equally favorable in both the forward and reverse directions, i.e., the concentrations of reactants and products are equal at equilibrium.

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 copper(I) ion forms a chloride salt that has \(K_{\mathrm{sp}}=1.2 \times\) \(10^{-6} .\) Copper(I) also forms a complex ion with \(\mathrm{Cl}^{-}:\) $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \quad K=8.7 \times 10^{4}$$ a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in \(0.10 M\) NaCl.

A friend tells you: "The constant \(K_{\mathrm{sp}}\) of a salt is called the solubility product constant and is calculated from the concentrations of ions in the solution. Thus, if salt A dissolves to a greater extent than salt \(\mathbf{B}\), salt \(\mathbf{A}\) must have a higher \(K_{\mathrm{sp}}\) than salt \(\mathbf{B}\)." Do you agree with your friend? Explain.

Use the following data to calculate the \(K_{\mathrm{sp}}\) value for each solid. a. The solubility of \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) is \(4.8 \times 10^{-5} \mathrm{mol} / \mathrm{L}\) b. The solubility of \(\mathrm{BiI}_{3}\) is \(1.32 \times 10^{-5} \mathrm{mol} / \mathrm{L}\)

Use the following data to calculate the \(K_{\mathrm{sp}}\) value for each solid. a. The solubility of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is \(6.2 \times 10^{-12} \mathrm{mol} / \mathrm{L}\). b. The solubility of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) is \(7.4 \times 10^{-2} \mathrm{mol} / \mathrm{L}\).

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties. a. \(\mathrm{Ag}_{3} \mathrm{PO}_{4}, K_{\mathrm{sp}}=1.8 \times 10^{-18}\) b. \(\mathrm{CaCO}_{3}, K_{\mathrm{sp}}=8.7 \times 10^{-9}\) c. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}, K_{\mathrm{sp}}=1.1 \times 10^{-18}\) \(\left(\mathrm{Hg}_{2}^{2+}\right.\)is the cation in solution.)
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

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