   Chapter 16.8, Problem 16.9CYU

Chapter
Section
Textbook Problem

What is the pH of a 0.10 M solution of oxalic acid, H2C2C4? What are the concentrations of H3O+, HC2O4−, and the oxalate ion, C2O42−? (See Appendix H for Ka values.)

Interpretation Introduction

Interpretation:

pH of 0.10 M solution of oxalic acid (H2C2O4) has to be calculated. The concentrations of H3O+, HC2O4- and C2O42- have to be calculated.

Concept introduction:

Dissociation constants are the equilibrium constants for the ionization reactions.

For the general acid HA:

HA (aq) + H2O (l) H3O+ (aq) + A-(aq)                 Ka = [H3O+][A-][HA]

For the general base,

B (aq) + H2O (l) BH+(aq) + OH-(aq)               Kb[BH+][OH-][B]

Explanation

From the given,

Concentration of oxalic acid = 0.10 M.

Oxalic acid is a diprotic acid.

The equilibrium reaction is as follows.

H2C2O4(aq) + H2O(l) H3O+(aq) + HC2O4-(aq)

The constant for the reaction is as follows.

Ka1 =  [H+][HC2O4-][H2C2O4]

The change that occurs as the reaction proceeds to equilibrium and the concentrations hen equilibrium has been established.

H2C2O4(aq)  +H2O(l) H3O+(aq) + HC2O4-(aq)I0.10                     0               0   C-x               x      xE(0.10-x)              x      x

Substitute the all values form the ICE table.

5.9×10-2 = (x)(x)[H2C2O4]5.9×10-2 = (x)2(0.10x)x=0.05278=5.28×102M=5.3×102M

Therefore, concentration of [H3O+] after the first ionization is 5.3×102M.

[H3O+]=[HC2O4]=5.3×102M

Let’s calculate the pH of the solution:

pH= -log[H3O+]= -log(5.3×102M)= 1

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