Chapter 17.2, Problem 17.4CYU

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Using an acetic acid/sodium acetate buffer solution, what ratio of conjugate base to acid will you need to maintain the pH at 5.00? Describe how you would prepare such a solution.

Interpretation Introduction

Interpretation:

Concept introduction:

The Henderson-Hasselbalch equation relates pH of a buffer with pKa of acid, concentration of conjugate base and concentration of acid. The expression is written as,

pH=pKa+log[conjugatebase][acid] (1)

This equation shows that pH of buffer solution is controlled by two major factors. First, Strength of the acid can be expressed on terms of pKa and second, the relative concentration of acid and its conjugate base at equilibrium. It can be seen from equation (1) that pH of the buffer solution is comparable to pKa values. So this equation can be used to establish a relation between pH and pKa value of acid.

Explanation

The ratio of acid to conjugate base by using Henderson-Hasselbalch equation is calculated below. The equilibrium between acetic acid and its conjugate base acetate is written as,

CH3COOH(aq)+H2O(l)â‡ŒH3O+(aq)â€‰+â€‰CH3COOâˆ’(aq)Â Â (acid)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (conjugateÂ base)

Therefore, acid is CH3COOH and its conjugate base is CH3COOâˆ’.

Given:

Refer to example number 17.4 in the text book for the value of Ka Â and pKa.

The value of Ka is 1.8Ã—10âˆ’5.

The value of pKa is 4.74.

The value of pH for the buffer solution to be prepared is 5.0.

Calculate the ratio of acid to conjugate base by using Henderson-Hasselbalch equation.

pH=pKa+log[conjugateâ€‰base][acid]

Substitute 4.74 for pKa and 5.0 for pH, [CH3COOH] for [acid] and [CH3COOâˆ’] for [conjugateâ€‰base],

5.0=4.74+log[CH3COOâˆ’][CH3COOH]

Rearrange for, log[CH3COOâˆ’][CH3COOH]

log[CH3COOâˆ’][CH3COOH]=5

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