   Chapter 16, Problem 109GQ

Chapter
Section
Textbook Problem

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

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

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