Chapter 17, Problem 16PS

### 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

# A buffer solution is composed of 1.360 g of KH2PO4 and 5.677 g of Na2HPO4. (a) What is the pH of the buffer solution? (b) What mass of KH2PO4 must be added to decrease the buffer solution pH by 0.50 unit from the value calculated in part (a)?

a)

Interpretation Introduction

Interpretation:

For the given buffer solution H2PO4/HPO42 of calculate (a) the value of when 1.360g of KH2PO4 and 5.677g of Na2HPO4 are mixed. (b) the mass of KH2PO4 must be added to decrease the  pH value of the buffer 0.50 unit from the value of pH in part (a).

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.

Explanation

The calculation of pH is done by using Henderson-Hasselbalch equation. The net ionic equilibrium between H2PO4âˆ’ and its conjugate base, HPO42âˆ’ is written as,

H2PO4âˆ’(aq)+H2O(l)â‡ŒH3O+(aq)â€‰+â€‰HPO42âˆ’(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 (conjugateÂ base)

Given:

Refer to table 16.2 in the textbook for the value of Ka.

The value of Ka for H2PO4âˆ’ 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 the H2PO4âˆ’ is 7.20.

The amount of KH2PO4 dissolved is 1.360â€‰g.

The amount of Na2HPO4 dissolved is 5.677â€‰g.

The molecular mass of KH2PO4 is 136.0â€‰gâ‹…molâˆ’1.

The molecular mass of Na2HPO4 is 141.96â€‰gâ‹…molâˆ’1.

The concentration of KH2PO4 and Na2HPO4 is calculated by using the following expression.

Molarity=weight(molecularâ€‰mass)(1â€‰Lâ€‰volumeâ€‰ofâ€‰solvent)(molâ‹…Lâˆ’1) (2)

Substitute, 1.360â€‰g for weight, 136.0â€‰gâ‹…molâˆ’1 for molecularâ€‰mass.

Molarity=1.360â€‰g(136.0â€‰gâ‹…molâˆ’1)(1â€‰Lâ€‰volumeâ€‰ofâ€‰solvent)(molâ‹…Lâˆ’1)=0

(b)

Interpretation Introduction

Interpretation:

For the given buffer solution H2PO4/HPO42 of calculate (a) the value of when 1.360g of KH2PO4 and 5.677g of Na2HPO4 are mixed. (b) the mass of KH2PO4 must be added to decrease the  pH value of the buffer 0.50 unit from the value of pH in part (a).

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.

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