   # Calculate the pH of a solution that is 0.40 M H 2 NNH 2 and 0.80 M H 2 NNH 3 NO 3 . In order for this buffer to have pH = p K a , would you add HCl or NaOH? What quantity (moles) of which reagent would you add to 1.0 L of the original buffer so that the resulting solution has pH = p K a ? ### 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 49E
Textbook Problem
352 views

## Calculate the pH of a solution that is 0.40 M H2NNH2 and 0.80 M H2NNH3NO3. In order for this buffer to have pH = pKa, would you add HCl or NaOH? What quantity (moles) of which reagent would you add to 1.0 L of the original buffer so that the resulting solution has pH = pKa?

Interpretation Introduction

Interpretation:

The concentration of H2NNH2 and H2NNH3NO3 is given. The pH of this solution is to be calculated. The compound to be added, HCl and NaOH , in order to have pH=pKa is to be identified. The quantity (moles) of reagent that would be added to 1.0L of the original buffer so that the resulting solution has pH=pKa is to be calculated.

Concept introduction:

A solution that contains mixture of a weak acid and its conjugate base is known as buffer solution. The pH value is the measure of H+ ions. The best buffer is prepared when the concentration of weak acid is equal to its conjugate base, that is [Acid]=[Base] .

To determine: The pH of solution of H2NNH2 and H2NNH3NO3 ; if HCl or NaOH is added in order to have pH=pKa ; the quantity (moles) of reagent that would be added to 1.0L of original buffer so that the resulting solution has pH=pKa

### Explanation of Solution

Explanation

To find the value of Ka

The value of Ka is calculated using the formula,

KaKb=Kw

Where,

• Kw is ion product constant (1.0×1014) .
• Kb is base dissociation constant.
• Ka is acid dissociation constant.

Substitute the values of Kb and Kw in the above equation.

KaKb=KwKa(3.0×106)=(1.0×1014)Ka=3.3×109_

To find the pH of solution of H2NNH2 and H2NNH3NO3

The concentration of H2NNH2 is 0.40M .

The concentration of H2NNH3NO3 is 0.80M .

The value of Ka of H2NNH2 is 3.3×109 .

Formula

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

pH=pKa+log[H2NNH2][H2NNH3+]

Where,

• pH is the measure of H+ ions.
• pKa is the measure of acidic strength.
• [H2NNH2] is concentration of (H2NNH2) .
• [H2NNH3+] is concentration of H2NNH3+ .

The formula of pKa is,

pKa=logKa

Where,

• Ka is acid equilibrium constant.

Substitute the value of pKa in Henderson-Hasselbach equation.

pH=logKa+log[H2NNH2][H2NNH3+]

Substitute all the given values in the above equation.

pH=logKa+log[H2NNH2][H2NNH3+]=log(3.3×109)+log[0.40][0.80]=8.18_

To find the base added to the buffer

In order to have pH=pKa , the concentration of weak acid is equal to its conjugate base, that is [Acid]=[Base] .

The value of Ka is 3.3×109 .

The formula of pKa is,

pKa=logKa

Where,

• Ka is acid equilibrium constant

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