   # Consider 1.0 L of a solution that is 0.85 M HOC 6 H 5 and 0.80 M NaOC 6 H 5 . ( K a for HOC 6 H 5 = 1.6 × 10 −10 .) a. Calculate the pH of this solution. b. Calculate the pH after 0.10 mole of HCl has been added to the original solution. Assume no volume change on addition of HCl. c. Calculate the pH after 0.20 mole of NaOH has been added to the original buffer solution. Assume no volume change on addition of NaOH. ### 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 14, Problem 100CWP
Textbook Problem
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## Consider 1.0 L of a solution that is 0.85 M HOC6H5 and 0.80 M NaOC6H5. (Ka for HOC6H5 = 1.6 × 10−10.)a. Calculate the pH of this solution.b. Calculate the pH after 0.10 mole of HCl has been added to the original solution. Assume no volume change on addition of HCl.c. Calculate the pH after 0.20 mole of NaOH has been added to the original buffer solution. Assume no volume change on addition of NaOH.

(a)

Interpretation Introduction

Interpretation: The pH of the given solution is to be calculated. The pH of the solution after the addition of 0.10 mol of HCl and 0.20 mol of NaOH is to be stated and no change in volume is to be assumed.

Concept introduction: The hydrogen ion concentration of the solution is known as pH of the solution.

It is the negative logarithm of Hydrogen ion concentration.

To determine: The pH of the given solution.

### Explanation of Solution

Explanation

Given

The given 1L solution is the combination of 0.85MC6H5OH and 0.80MC6H5ONa.

The value of equilibrium constant or acid dissociation constant (Ka) for C6H5OH is 1.6×1010.

The given solution is the combination of acid and their salt. Therefore the pH of such type of solution is calculated by the equation,

pH=pKa+log[salt][acid]

Where,

• pKa is the negative logarithm of dissociation constant.
• [salt] is the concentration of salt.
• [acid] is the concentration of the acid.

The above equation is known as Henderson-Hasselbalch equation.

The pKa that is a acid dissociation constant is determined by the formula.

pKa=log(Ka)

Substitute the given value of Ka in the above equation

(b)

Interpretation Introduction

Interpretation: The pH of the given solution is to be calculated. The pH of the solution after the addition of 0.10mol of HCl and 0.20mol of NaOH is to be stated and no change in volume is to be assumed.

Concept introduction: The hydrogen ion concentration of the solution is known as pH of the solution.

It is the negative logarithm of Hydrogen ion concentration.

To determine: The value of pH after the addition of 0.10mol of HCl in the given solution.

(c)

Interpretation Introduction

Interpretation: The pH of the given solution is to be calculated. The pH of the solution after the addition of 0.10mol of HCl and 0.20mol of NaOH is to be stated and no change in volume is to be assumed.

Concept introduction: The hydrogen ion concentration of the solution is known as pH of the solution.

It is the negative logarithm of Hydrogen ion concentration.

To determine: The value of pH after the addition of 0.20mol of NaOH in the given solution.

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