   # Calculate the pH of a solution that contains 1.0 M HF and 1.0 M HOC 6 H 5 . Also calculate the concentration of OC 6 H 5 − in this solution at equilibrium. ### 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 71E
Textbook Problem
114 views

## Calculate the pH of a solution that contains 1.0 M HF and 1.0 M HOC6H5. Also calculate the concentration of OC6H5− in this solution at equilibrium.

Interpretation Introduction

Interpretation: The concentration of OC6H5 and the pH of the given solution, containing 1.0M HF and 1.0M HOC6H5 , is to be calculated.

Concept introduction: The pH of a solution is define as a figure that expresses the acidity of the alkalinity of a given solution. A logarithmic scale is used on which, the value 7 corresponds to a neutral species, a value less than 7 corresponds to an acid and a value greater than 7 corresponds to a base.

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

The equilibrium constant for water is denoted by Kw and is expressed as,

Kw=[H+][OH]

### Explanation of Solution

Explanation

To determine: The concentration of OC6H5 and the pH of given solution.

The [H+] is 0.026M_ .

In the given solution,

The Ka for HF is 7.2×104 .

The Ka for C6H5OH is 1.6×1010 .

The Ka of HF is greater than that of C6H5OH . Therefore, HF is a strong acid.

The dominant equilibrium reaction for the given case is,

HF(aq)H+(aq)+F(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+][F][HF] (1)

The [H+] is 2.6×10-2M_ .

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

The ICE table for the stated reaction is,

HF(aq)H+(aq)+F(aq)Inititialconcentration1.000Changex+x+xEquilibriumconcentration1.0xxx

The equilibrium concentration of [HF] is (1.0x)M .

The equilibrium concentration of [H+] is xM .

The equilibrium concentration of [F] is xM .

The Ka for HF is 7.2×104 .

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

7.2×104=[x][x][1.0x]7.2×104=[x]2[1.0x]

The value of x will be very small as compared to 1.0 . Hence, it is ignored from the term [1.0x] .

Simplify the above expression.

7.2×104=[x]2[1.0][x]2=(7

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