   # Consider the titration of 150.0 mL of 0.100 M HI by 0.250 M NaOH. a. Calculate the pH after 20.0 mL of NaOH has been added. b. What volume of NaOH must be added so that the pH = 7.00? ### 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 104CWP
Textbook Problem
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## Consider the titration of 150.0 mL of 0.100 M HI by 0.250 M NaOH.a. Calculate the pH after 20.0 mL of NaOH has been added.b. What volume of NaOH must be added so that the pH = 7.00?

(a)

Interpretation Introduction

Interpretation: The pH of the solution after the addition of 20.00mL of 0.250M NaOH is to be calculated. To make the pH7.00 volume of the NaOH is to be calculated..

Concept introduction: HI is a strong acid and NaOH is a strong base. The reaction between an acid and a base takes place with the formation of a salt and a water molecule.

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

It is the negative logarithm of Hydrogen ion concentration.

The number of moles calculated by taking the volume in milliliter is known as number of milli moles.

pH-7 is the neutralization condition of every solution.

To determine: The pH of the solution after the addition of 20.00mL of NaOH.

### Explanation of Solution

Explanation

Given

Titration of 150.0ml of 0.100MHI with 0.250MNaOH.

The number of milli moles is calculated by the formula,

n=C×V

Where,

• n is the number of milli moles.
• C is the concentration of the solution.
• V is the volume of the solution.

Substitute the values of concentrations and volumes in the above equation.

The number of mill moles of NaOH,

n=C×V=20.0ml×0.250M=5.00mmol

The number of mill moles of HI,

n=C×V=150.0ml×0.100M=15.0mmol

The calculated number of milli moles states that 5mmol of NaOH neutralize only 5mmol of HI. Therefore, the number of remaining mmol is of HI only and 155=10mmol

(b)

Interpretation Introduction

Interpretation: The pH of the solution after the addition of 20.00mL of 0.250M NaOH is to be calculated. To make the pH7.00 volume of the NaOH is to be calculated..

Concept introduction: HI is a strong acid and NaOH is a strong base. The reaction between an acid and a base takes place with the formation of a salt and a water molecule.

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

It is the negative logarithm of Hydrogen ion concentration.

The number of moles calculated by taking the volume in milliliter is known as number of milli moles.

pH-7 is the neutralization condition of every solution.

To determine: The volume of the NaOH added to make the pH-7 in the given solution.

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