   # Calculate the pH after 0.10 mole of NaOH is added to 1.00 L of the solution in Exercise 31, and calculate the pH after 0.20 mole of HCl is added to 1.00 L of the solution in Exercise 31. ### 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 33E
Textbook Problem
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## Calculate the pH after 0.10 mole of NaOH is added to 1.00 L of the solution in Exercise 31, and calculate the pH after 0.20 mole of HCl is added to 1.00 L of the solution in Exercise 31.

Interpretation Introduction

Interpretation:

The pH value after 0.10mol NaOH is added to 1L of the solution in Exercise 31 and the pH value after 0.20mol HCl is added to 1L of the solution in exercise 31 is to be calculated.

Concept introduction:

A solution that contains a mixture of a weak acid and its conjugate base is known as a buffer solution.

The pH value is the measure of H+ ions. The pH is calculated by the formula,

pH=log10[H+]

To determine: The pH value after 0.10mol NaOH is added to 1L of the solution in exercise 31 ; the pH value after 0.20mol HCl is added to 1L of the solution in exercise 31 .

### Explanation of Solution

Explanation

Given

Volume of solution is 1.0L .

Number of moles of NaOH added is 0.10mol .

Refer Exercise 31 .

The pH value of the solution of 1.00M HNO2 and 1.00M NaNO2 is 3.40 .

The concentration of HNO2 is 1.00M .

The concentration of NaNO2 is 1.00M .

Formula

The number of moles of compound in a solution is calculated by the formula,

Molesofcompound=Concentrationofcompound×Volumeofsolution (1)

Substitute the values of concentration and volume of HNO2 in the above equation.

Molesofcompound=Concentrationofcompound×Volumeofsolution=1.0M×1.0L=1.0mol_

Substitute the values of concentration and volume of NaNO2 in equation (1).

Molesofcompound=Concentrationofcompound×Volumeofsolution=1.0M×1.0L=1.0mol_

To determine the final number of moles of HNO2 and NaNO2

The dissociation reaction of HNO2 is,

HNO2+OHNO2+H2O

Initial moles of HNO2 is 1.00mol .

On addition of the given concentration of NaOH to the buffer solution, the OH ions react with HNO2 and lead to the formation of NO2 ions.

The final moles of NO2 are,

1.00mol+0.10mol=1.10mol_

The final moles of HNO2 are,

1.00mol0.10mol=0.90mol_

To determine the value of pKa

The value of Ka of HNO2 is 4.3×104 .

The formula of pKa is,

pKa=logKa

Where,

• Ka is acid equilibrium constant.

Substitute the value of Ka in the above equation.

pKa=logKa=log(4.3×104)=3.37_

To determine the   pH

As the volume of solution is 1.0L , the concentration of compound is equal to number of moles.

Therefore, the concentration of HNO2 is 0.90M and of NO2 is 1.10M

The value of pKa is 3.37 .

Formula

The pH is calculated using the Henderson-Hassel Bach equation,

pH=pKa+log[NO2][HNO2]

Where,

• pH is the measure of H+ ions.
• pKa is the measure of acidic strength.
• [NO2] is concentration of NO2

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