   # You prepare a 0.10 M solution of oxalic acid, H 2 C 2 O 4 . What molecules and ions exist in this solution? List them in order of decreasing concentration. ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 16, Problem 107GQ
Textbook Problem
6 views

## You prepare a 0.10 M solution of oxalic acid, H2C2O4. What molecules and ions exist in this solution? List them in order of decreasing concentration.

Interpretation Introduction

Interpretation:

The decreasing order of the ions concetration in the ionization of oxalic acid has to be determined.

Concept introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka is an acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb is a base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

### Explanation of Solution

The oxalic acid is a diprotic acid. It ionizes into two steps.

H2C2O4(aq) + H2O(l)HC2O4-(aq) + H3O+(aq)    Ka1=5.9×10-2HC2O4-(aq) + H2O( l)C2O4-(aq) + H3O+(aq)       Ka2=6.4×10-5

From the ionization steps,

0.10 M solution contains the following ions.

H2C2O4, H2O, HC2O4-, C2O42-, H3O+and OH-

Let’s calculate the ions concentration by using ICE table:

For the first ionization of oxalic acid:

Equilibrium        H2C2O4(aq) + H2O(l) HC2O4-(aq) + H3O+(aq)I                                0.10             --                 0                   0C                                -x               --                 +x                +xE                             (0.10-x)        --                   x                   x

The equilibrium expression:

Ka1=[HC2O4-][H3O+][H2C2O4]

Substitute the values from ICE table

5.9×10-2=(x)(x)(0.10-x)

Here, [H2C2O4]initial< 100Ka1(0.10 < 5.9)

x2= (5.9×10-2)(0.10-x)x2= 5.9×10-3-5.9×10-2-x

All variants are equal to zero.

Therefore,

x2+5.9×10-2 x-5.9×10-3= 0

Solve the quadratic equation, we get x value.

x =-5.9×10-2±(5.9×10-2)2-4(1)(-5.9×10-3)2(1)x =-5.9×10-2±34.81×10-4+23.6×10-32x = 0

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts
What are the moral implications of using ergogenic aids?

Understanding Nutrition (MindTap Course List)

An important national food and nutrient intake survey, called What We Eat in America, is part of a. NHANES. b. ...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

Define the following terms: a. chromosome b. chromatin

Human Heredity: Principles and Issues (MindTap Course List)

How many significant figures does 0.00130 m have?

Physics for Scientists and Engineers: Foundations and Connections

How are wind waves formed? Whats a fetch?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin

A 70.0-kg log falls from a height of 25.0 m into a lake. If the log, the lake, and the air are all at 300 K, fi...

Physics for Scientists and Engineers, Technology Update (No access codes included) 