Chapter 17, Problem 20PS

### 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 NH3 and NH4Cl to have a pH of 9.5.

Interpretation Introduction

Interpretation:

To make the buffer solution contains NH3 and NH4+ and its pH value is 9.5 has to be explained.

Concept introduction:

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

pOH=pKb+log[conjugateacid][base] (1)

This equation shows that pOH of buffer solution is controlled by two major factors. First, is strength of the base which can be expressed on terms of pKb and second, the relative concentration of base and its conjugate acid at equilibrium.

Explanation

The equilibrium between â€‰NH3 and its conjugate base â€‰NH4+ is written as follows;

NH3(aq)+H2O(l)â‡ŒNH4+(aq)+â€‰OHâˆ’(aq)â€‰â€‰(base)â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰(conjugateâ€‰acid)

The value of pOH is calculated by using expression,pHÂ +Â pOHÂ =Â 14.

Given:

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

The value of Kb for â€‰NH3 is 1.8Ã—10âˆ’5.

Negative logarithm of the Kb value gives the pKb value of the acid.

pKb=âˆ’log(Kb)=âˆ’log(1.8Ã—10âˆ’5)=4.74

Therefore, pKb value for the â€‰NH3 is 4.74.

The pH for the given buffer solution is 9.5.

The value of pOH is calculated by using expression,

pHÂ +Â pOHÂ =Â 14

Rearrange for pOH,

pOHÂ =Â 14âˆ’pHÂ Â

Substitute, 9.5 for pH,

pOHÂ =Â 14âˆ’9.5=â€‰4.5

Therefore, pOH value for the given buffer is 4.5.

The value of concentration ratio of base and its conjugate acid is calculated by using equation (1).

pOH=pKb+log[conjugateâ€‰acid][base]

Substitute, 4.74 for pKb, 4.5 for pOH, [NH4+]Â  for [conjugateÂ acid], [NH3] for

[base]

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