   # Calculate the pH and [S 2− ] in a 0.10- M H 2 S solution. Assume K a 1 = 1.0 × 10 −7 ; K a 2 = 1.0 × 10 −19 . ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 13, Problem 109E
Textbook Problem
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## Calculate the pH and [S2−] in a 0.10-M H2S solution. Assume K a 1 = 1.0 × 10−7; K a 2 = 1.0 × 10−19.

Interpretation Introduction

Interpretation: The pH of the given H2S solution and the [S2] is to be calculated.

Concept introduction: The pH of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The pH of a solution is calculated by the formula, pH=log[H+]

At equilibrium, the equilibrium constant expression is expressed by the formula,

Ka=ConcentrationofproductsConcentrationofreactants

### Explanation of Solution

Explanation

To determine: T The pH of the given H2S solution and the [S2] .

The equilibrium constant expression for the initial dissociation reaction is,

Ka=[H+][HS][H2S]

The initial dissociation reaction is,

H2S(aq)H+(aq)+HS(aq)

At equilibrium, the equilibrium constant expression is expressed by the formula,

Ka=ConcentrationofproductsConcentrationofreactants

Where,

• Ka is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

Ka=[H+][HS][H2S] (1)

The [H+] from first reaction is 1.0×10-4M_ .

The change in concentration of H2S is assumed to be x .

The ICE table for the stated reaction is,

H2S(aq)H+(aq)+HS(aq)Inititialconcentration0.1000Changex+x+xEquilibriumconcentration0.10xxx

The equilibrium concentration of [H2S] is (0.10x)M .

The equilibrium concentration of [H+] is xM .

The equilibrium concentration of [HS] is xM .

The Ka for the first reaction is given to be is 1.0×107 .

Substitute the value of Ka , [H2S] , [H+] and [HS] in equation (1).

1.0×107=[x][x][0.10x]1.0×107=[x]2[0.10x]

The value of x will be very small as compared to 0.10 . Hence, it is ignored from the term [0.10x] .

Simplify the above expression.

1.0×107=[x]2[0.10][x]2=(1.0×108)[x]=1.0×10-4M_

Therefore, the [H+] from the first reaction is 1.0×10-4M_ .

The initial [HS] is 1.0×10-4M_ .

According to the ICE table formed,

The [HS] is equal to the [H+] ,that is 1.0×104M .

This is the initial [HS] for the further reaction.

The [H+] from second reaction is 1.0×10-19M_ .

The change in concentration of HS is assumed to be y .

The ICE table for the stated reaction is,

HS(aq)H+(aq)+S2(aq)Inititial1.0×1041.0×1040Changey+y+yEquilibrium(1

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