Introduction to General, Organic and Biochemistry
Introduction to General, Organic and Biochemistry
11th Edition
ISBN: 9781285869759
Author: Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher: Cengage Learning
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Textbook Question
Chapter 8, Problem 8.101P

8-101 Suppose you have an aqueous solution prepared by dissolving 0.050 mol of NaH2PO4 in 1 L of water. This solution is not a buffer, but suppose you want to make it into one. How many moles of solid Na2HPO4 must you add to this aqueous solution to make it into:

(a) A buffer of pH 7.21

(b) A buffer of pH 6.21

(c) A buffer of pH 8.21

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

The number of moles to be added in NaH2 PO4 to make it into a buffer of pH 6.21 should be calculated.

Concept Introduction:

The pH of buffer solution is calculated using the following formula:

 pH = pKa+ log[A][HA]

This is known as Henderson- Hasselbalch equation. Here, pKa is negative log of acid dissociation constant, [A] is concentration of conjugate base and [HA] is concentration of acid.

Answer to Problem 8.101P

0.050 mole of NaH2PO4.

Explanation of Solution

Given Information:

An aqueous solution contains 0.050 moles of NaH2 PO4.

The volume of the solution is 1 L.

The pH of the buffer solution is 7.21.

A buffer is a solution which resists change in pH when limited amounts of an acid or a base are added to it. The given buffer is made of the weak acid, NaH2 PO4 and its conjugate base Na2 HPO4.

The Henderson-Hasselbalch equation for a buffer of a weak acid (HA) and its conjugate base(A- ) is given as:

pH = pKa+ log[A][HA]

By using the Henderson-Hasselbalch equation for the pH of the buffer solution we get,

pH = pKa+ log[Na2HPO4][NaH2PO4]

As, we know that pKa of NaH2 PO4 is 7.21.

The volume of the buffer solution is 1 L. Assuming that there will not be any change in the volume of solution due to the addition of Na2 HPO4, the concentration terms in the Henderson-Hasselbalch equation can be replaced by the number of moles.

By substituting the values of pH, pKa and the number of moles of NaH2 PO4 for [NaH2 PO4 ] in the above equation, we get.

7.21 = 7.21+ log[Na2HPO4]0.0500=log[Na2HPO4]0.050

Bu taking antilog on both sides we get,

antilog(0) = [Na2HPO4]0.050             1 = [Na2HPO4]0.050[Na2HPO4] = 0.050

Thus, 0.050 moles of Na2 HPO4 must be added to the solution of NaH2 PO4 to make a buffer of 7.21. This also shows that when the equimolar of a weak acid and its conjugate base are taken, the pH of the buffer is equal to the pKa of the weak acid.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

The number of moles to be added in NaH2 PO4 to make it into a buffer of pH 7.21 should be calculated.

Concept Introduction:

The pH of buffer solution is calculated using the following formula:

 pH = pKa+ log[A][HA]

This is known as Henderson- Hasselbalch equation. Here, pKa is negative log of acid dissociation constant, [A] is concentration of conjugate base and [HA] is concentration of acid.

Answer to Problem 8.101P

0.005 mole of NaH2PO4.

Explanation of Solution

Given Information:

An aqueous solution contains 0.050 moles of NaH2 PO4.

The volume of the solution is 1 L.

The pH of the buffer solution is 6.21.

By using the Henderson-Hasselbalch equation for the pH of the buffer solution we get,

pH = pKa+ log[Na2HPO4][NaH2PO4]

As, we know that pKa of NaH2 PO4 is 7.21.

The volume of the buffer solution is 1 L. Assuming that there will not be any change in the volume of solution due to the addition of Na2 HPO4, the concentration terms in the Henderson-Hasselbalch equation can be replaced by the number of moles.

By substituting the values of pH, pKa and the number of moles of NaH2 PO4 for [NaH2 PO4 ] in the above equation, we get.

6.21 = 7.21+ log[Na2HPO4]0.0501=log[Na2HPO4]0.050

Bu taking antilog on both sides we get,

antilog(-1.00) = [Na2HPO4]0.050                    1 = [Na2HPO4]0.050  [Na2HPO4] = 0.005

Thus, 0.005 moles of Na2 HPO4 must be added to the solution of NaH2 PO4 to make a buffer of 6.21.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

The number of moles to be added in NaH2 PO4 to make it into a buffer of pH 8.21 should be calculated.

Concept Introduction:

The pH of buffer solution is calculated using the following formula:

 pH = pKa+ log[A][HA]

This is known as Henderson- Hasselbalch equation. Here, pKa is negative log of acid dissociation constant, [A] is concentration of conjugate base and [HA] is concentration of acid.

Answer to Problem 8.101P

0.500 mole of Na2HPO4.

Explanation of Solution

Given Information:

An aqueous solution contains 0.050 moles of NaH2 PO4.

The volume of the solution is 1 L.

The pH of the buffer solution is 8.21.

By using the Henderson-Hasselbalch equation for the pH of the buffer solution we get,

pH = pKa+ log[Na2HPO4][NaH2PO4]

As, we know that pKa of NaH2 PO4 is 7.21.

The volume of the buffer solution is 1 L. Assuming that there will not be any change in the volume of solution due to the addition of Na2 HPO4, the concentration terms in the Henderson-Hasselbalch equation can be replaced by the number of moles.

By substituting the values of pH, pKa and the number of moles of NaH2 PO4 for [NaH2 PO4 ] in the above equation, we get.

8.21 = 7.21+ log[Na2HPO4]0.0501.00=log[Na2HPO4]0.050

Bu taking antilog on both sides we get,

antilog(1.00) = [Na2HPO4]0.050                 10 = [Na2HPO4]0.050 [Na2HPO4] = 0.500

Thus, 0.500 moles of Na2 HPO4 must be added to the solution of NaH2 PO4 to make a buffer of 8.21.

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Chapter 8 Solutions

Introduction to General, Organic and Biochemistry

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