   # What volumes of 0.50 M HNO 2 and 0.50 M NaNO 2 must be mixed to prepare 1.00 L of a solution buffered at pH = 3.55? ### 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 42E
Textbook Problem
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## What volumes of 0.50 M HNO2 and 0.50 M NaNO2 must be mixed to prepare 1.00 L of a solution buffered at pH = 3.55?

Interpretation Introduction

Interpretation:

The concentration of HNO2,NaNO2 , volume and the required pH of the buffer solution is given. The volume of HNO2,NaNO2 that must be added to make the buffer solution is to be calculated.

Concept introduction:

The pH value is the measure of H+ ions. The relation between pH and pKa is given by Henderson-Hassel Bach equation. According to this equation,

pH=pKa+log[Base][Acid]

### Explanation of Solution

Explanation

To find the ratio, [NaNO2][HNO2]

Given

The concentration of HNO2 is 0.50M .

The concentration of NaNO2 is 0.50M .

The value of Ka for HNO2 is 4.0×104 .

The pH value of solution is 3.55 .

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

pH=pKa+log[NaNO2][HNO2]

Where,

• pH is the measure of H+ ions.
• pKa is the measure of acidic strength.
• [NaNO2] is the concentration of NaNO2 .
• [HNO2] is the concentration of HNO2 .

The formula of pKa is,

pKa=logKa

Where,

• Ka is acid equilibrium constant.

Substitute the value of pKa in the above equation.

pH=logKa+log[NaNO2][HNO2]

Substitute all the given values in the above equation.

pH=logKa+log[NaNO2][HNO2]3.55=log(4.0×104)+log[NaNO2][HNO2][NaNO2][HNO2]=1.42_ (1)

To find the volume of HNO2

It is assumed that volume of NaNO2 is x and volume of HNO2 is y

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