Chapter 17, Problem 19PS

### 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

# Describe how to prepare a buffer solution from NaH2PO4 and Na2HPO4 to have a pH of 7.5.

Interpretation Introduction

Interpretation:

The concentration of acid and its base has to be calculated to make the buffer solution of pH value equals to 7.5.

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 calculation of the ratio of concentration of acid to conjugate base is calculated by using Henderson-Hasselbalch equation. The equilibrium between NaH2PO4â€‰andâ€‰Na2HPO4 is written as follows.

â€‚Â NaH2PO4(aq)+H2O(l)â‡ŒH3O+(aq)â€‰+â€‰Na2HPO4(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 (conjugateÂ base)

Given:

Refer to table 16.2 in the text book for the values of Ka.

The value of Ka for NaH2PO4 is 6.2Ã—10âˆ’8.

Negative logarithm of the Ka value gives the pKa value of the acid.

pKa=âˆ’log(Ka)=âˆ’log(6.2Ã—10âˆ’8)=7.20

Therefore, pKa value for NaH2PO4 is 7.20.

The value of pH for the buffer solution which is to prepared solution is 7.5.

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

pH=pKa+log[conjugateâ€‰base][acid]

Substitute 7.20 for pKa and 7.5 for pH, [NaH2PO4] for [acid] and [Na2HPO4] for [conjugateâ€‰base],

7.5=7.20+log[Na2HPO4][NaH2PO4]

Rearrange for, log[Na2HPO4][NaH2PO4]

log[Na2HPO4][NaH2PO4]=7

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