   Chapter 16, Problem 71PS

Chapter
Section
Textbook Problem

Prove that Ka1 × Kb2 = Kw for oxalic acid H2C2O4, by adding the chemical equilibrium expressions that corresponds to first ionization step of the acid in water with second step of the reaction of the fully deprotonated base, C2O42−, with water.

Interpretation Introduction

Interpretation:

Ka1×Kb2=Kw has to be proved for oxalic acid H2C2O4 by adding the chemical equilibrium expressions that corresponds to first ionization step of the acid in water with second step of the reaction of the fully deprotonated base, C2O42-,with water.

Concept introduction:

If a molecule donate only one hydrogen atom then the acid is called a monoprotic acid.

Example: HCl, HBr and HClO3

If each molecule can donate two or more hydrogen ions, the acid is a polyprotic acid.

A diprotic acid, such as sulfuric acid, H2SO4 has two acidic hydrogen atoms.

Some acids, such as phosphoric acid, H3PO4 are triprotic acids.

Polyprotic bases:

Can accept more than on proton (H+). So, it can react with water many times to produce many hydroxide ions (OH).

Diprotic bases:

Can accept two protons.

Example: Sulphate ion SO4(aq)2-

Explanation

Oxalic acid H2C2O4 is a diprotic acid. It donates two hydrogen atoms.

Chemical equation for the first step of oxalic acid H2C2O4 is

First Ionization:

H2C2O4+H2OH3O+ + HC2O4-

Equilibrium constant:

Ka1[H+][HC2O4-][H2C2O4]

Chemical equation for the second step of oxalic acid H2C2O4 is

Second Ionization:

HC2O4-+H2OOH- +H2C2O4

Equilibrium constant:

Kb2

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