   Chapter 16, Problem 72PS

Chapter
Section
Textbook Problem

Prove that Ka3 × Kb1 = Kw for phosphoric acid, H3PO4, by adding the chemical equilibrium expressions that corresponds to the third ionization step of the acid in water with the first of the three successive steps of the reaction of phosphate ion, PO43−, with water.

Interpretation Introduction

Interpretation:

Ka3×Kb1= Kw has to be proved for phosphoric acid, H3PO4 by adding the chemical equilibrium expressions that corresponds to the third ionization step of the acid in water with the first of the three successive steps of the reaction of phosphate ion, PO43- 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

Phosphoric acid is a triprotic acid it donates three protons.

First ionization step of phosphoric acid:

H3PO4(aq) + H2O (l)H3O+(aq) +H2PO4-Ka1[H3O+][H2PO4-][H3PO4]

Second ionization

H2PO4-(aq) + H2O(l)H3O+(aq) + HPO42-(aq)Ka2[H3O+][H2PO42-][H2PO-4]

Third ionization

HPO42-(aq) +H2O(l)H3O+(aq) + PO43-(aq)K

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