   Chapter 15, Problem 48E

Chapter
Section
Textbook Problem

# Consider the bases in Table 13-3. Which base would be the best choice for preparing a pH = 5.00 buffer? Explain how to make 1.0 L of this buffer.

Interpretation Introduction

Interpretation:

The base that would be the best choice for preparing a pH=5.00 buffer is to be identified from the Table 13-3 . An explanation regarding the making 1.0L of this buffer.

Concept introduction:

A solution that contains mixture of a weak acid and its conjugate base is known as buffer solution. The pH value is the measure of H+ ions. The best buffer is prepared when the concentration of weak acid is equal to its conjugate base, that is [Acid]=[Base] .

To determine: The base that would be best choice for preparing a pH=5.00 buffer; the explanation to prepare 1.0L of this buffer.

Explanation

Explanation

To find the base, that best choice for a pH 5.00 buffer with Kb is 1.7×109 .

The pH is calculated using the Henderson-Hassel Bach equation,

pH=pKa+log[Base][Acid]

Where,

• pH is the measure of H+ ions.
• pKa is the measure of acidic strength.
• [Base] is concentration of conjugate base.
• [Acid] is concentration of weak acid.

Since, the best buffer is prepared when the concentration of weak acid is equal to its conjugate base, the above equation becomes,

pH=pKa+log[Base][Acid]pH=pKa+0pHpKa

The base with pKa close to 5.00 or Ka1×105 would be the best choice for a pH 5.00 buffer.

The value of Kb is calculated using the formula,

KaKb=Kw

Where,

• Kw is ion product constant (1.0×1014) .
• Kb is base dissociation constant.
• Ka is acid dissociation constant.

Substitute the values of Ka and Kw in the above equation.

KaKb=Kw1×105Kb=(1.0×1014)Kb=1×109

Hence, from the table 13-3 , Pyridine is the best choice for a pH 5.00 buffer with Kb is 1.7×109 .

To find Ka

The value of Kb for pyridine is 1.7×109 .

The value of pH is 5.00 .

The value of Ka is calculated using the formula,

KaKb=Kw

Where,

• Kw is ion product constant (1.0×1014) .
• Kb is base dissociation constant.
• Ka is acid dissociation constant.

Substitute the values of Kb and Kw in the above equation.

KaKb=KwKa(1.7×109)=(1.0×1014)Ka=5.9×106_

To find the ratio [C5H5N][C5H5NH+]

Given,

The value of Ka is (5

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