   Chapter 16, Problem 73PS

Chapter
Section
Textbook Problem

Sulphurous acid, H2SO3, is a weak acid capable of providing two H+ ions. (a) What is the pH of a 0.45M solution of H2SO3? (b) What is the equilibrium concentration of the sulphate ion SO32−, in the 0.45M solution of H2SO3?

(a)

Interpretation Introduction

Interpretation:

The pH of a 0.45M solution of H2SO3 has to be determined

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

Sulfurous acid is a diprotic acid.

First ionization:H2SO3(aq) + H2O(aq)H3O+(aq) + HSO3-(aq)Equilibrium expression:Ka1 = [HSO3-][H3O+][H2SO3]Ka1= 1.2×10-2Second ionization:HSO3-(aq) + H2O(aq)H3O+(aq) + SO32-(aq)Equilibrium expression:Ka2[H3O+][SO32-][HSO3-]Ka2= 6.2×10-8

From the Ka1 and Ka2 values,  Ka2 is smaller than the  Ka1.

Therefore, H3O+ is almost produced entirely from  Ka1.

Let’s calculate the H3O+ from  Ka1.

First ionization:H2SO3(aq) + H2O(aq)H3O+(aq) + HSO3-(aq)Equilibrium expression:Ka1 = [HSO3-][H3O+][H2SO3]Ka1= 1.2×10-2

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.

H2SO3(aq) + H2O (aq) H3O+(aq) + HSO3-(aq)I         0

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentration has to be determined for the sulphate ion SO32- in the 0.45M solution of H2SO3.

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 acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb is base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

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-

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