   # 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. ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 17.2, Problem 3CYU
Textbook Problem
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## 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 of Solution

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)H3O+(aq)+CH3COO(aq)  (acid)                                                     (conjugate base)

Therefore, acid is CH3COOH and its conjugate base is CH3COO.

Given:

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

The value of Ka is 1.8×105.

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[conjugatebase][acid]

Substitute 4.74 for pKa and 5.0 for pH, [CH3COOH] for [acid] and [CH3COO] for [conjugatebase],

5.0=4.74+log[CH3COO][CH3COOH]

Rearrange for, log[CH3COO][CH3COOH]

log[CH3COO][CH3COOH]=5

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