   # Calculate the pH of a solution that is 0.20 M HOCl and 0.90 M KOCl. 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 50E
Textbook Problem
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## Calculate the pH of a solution that is 0.20 M HOCl and 0.90 M KOCl. 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 HOCl and KOCl is given. The pH value of this solution is to be calculated. The compound to be added, between 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 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 HOCl and KOCl ; 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.

### Explanation of Solution

Explanation

To find the pH of solution of HOCl and KOCl

The concentration of HOCl is 0.20M .

The concentration of KOCl is 0.90M .

The value of Ka for HOCl is 3.5×108 .

Formula

The pH is calculated using the Henderson-Hasselbach equation,

pH=pKa+log[OCl][HOCl]

Where,

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

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[OCl][HOCl]

Substitute all the given values in Henderson-Hasselbach equation.

pH=logKa+log[OCl][HOCl]=log(3.5×108)+log[0.90][0.20]=8.11_

To find the acid that is needed to be added

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

But the concentration of acid is less than the concentration of base. Hence, in order to have [Acid]=[Base] , the concentration of acid should be increased. This can only happen when more acid is added to the solution of HOCl and KOCl . Therefore, HCl must be added to the buffer solution.

To show that the resulting solution has pH=pKa

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