Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Chapter 16, Problem 73PS

Sulphurous acid, H2SO3, is a weak acid capable of providing two H+ ions.

  1. (a) What is the pH of a 0.45M solution of H2SO3?
  2. (b) What is the equilibrium concentration of the sulphate ion SO32−, in the 0.45M solution of H2SO3?

(a)

Expert Solution
Check Mark
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-

Answer to Problem 73PS

pH of a 0.45M solution of H2SO3 is 1.17

Explanation of Solution

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         0.45M          --                       --                --C           -x              --                     +x              +xE        (0.45-x)       --                        x               x

Ka1 = [HSO3-][H3O+][H2SO3]Ka1= 1.2×10-2

At equilibrium,x = [H3O+] = [HSO3-]1.2×10-2 = x2(0.45-x)x2 + 1.2×10-2 - 5.4×10-3 = 0x = -1.2×10-2±(1.2×10-2)2-4(1)(-5.4×10-3)2(1)x = 6.77×10-2Therefore,[H3O+] = [HSO3-] = 6.77×10-2pH = -log[H3O+]pH = -log(6.77×10-2)      =1.17

(b)

Expert Solution
Check Mark
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-

Answer to Problem 73PS

The equilibrium concentration of the sulphate ion SO32- in the 0.45M solution of H2SO3 is 6.28×10-8M

Explanation of Solution

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

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.

        HSO3(aq)      + H2O (aq) SO32(aq) + H3O+(aq)I         6.77×102M          --                       --          6.77×102MC           -y                        --                     +y              +yE        (6.77×102-x)       --                        y          (6.77×102M)

Ka2[H3O+][SO32-][HSO3-]Ka2= 6.2×10-8

6.2×10-8 = y (6.77×10-2 + y)(6.77×10-2- y)Y is so small compared to 6.77×10-2So, all values are almost equal to 6.77×10-26.2×10-8 = y(6.77×10-2)(6.77×10-2)             y = 6.2×10-8

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Chapter 16 Solutions

Chemistry & Chemical Reactivity

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