# 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

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
1 views

## 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

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts
A represents a quantitative observation.

Introductory Chemistry: A Foundation

Recommendations about carbohydrate intake can seem to be contradictory. On one hand, it is recommended that the...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

What is genomics?

Human Heredity: Principles and Issues (MindTap Course List)

If the holdfast of this alga breaks free in a storm, will the organism die?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin

Why is the following situation impossible? Starting from rest, a disk rotates around a fixed axis through an an...

Physics for Scientists and Engineers, Technology Update (No access codes included)