   Chapter 16, Problem 16.122QP General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343

Solutions

Chapter
Section General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343
Textbook Problem

The equilibrium equations and Ka values for three reaction systems are given below. NH 4 + ( a q ) + H 2 O ⇌ H 3 O + + NH 3 ( a q ) ; K a = 5.6 × 10 − 10 H 2 CO 3 ( a q ) + H 2 O ⇌ H 3 O + ( a q ) + HCO 3 − ( a q ) ; K a = 4.3 × 10 − 7 H 3 PO 4 − ( a q ) + H 2 O ⇌ H 3 O + ( a q ) + HPO 4 2 − ( a q ) ; K a = 6.2 × 10 − 8 a Which conjugate pair would be best for preparing a buffer with a pH of 6.96? Why? b How would you prepare 100 mL of a buffer with a pH of 6.96 assuming that you had available 0.10 M solutions of each pair?

(a)

Interpretation Introduction

Interpretation:

From the given equilibrium equations and Ka value for the given three reaction systems,

The conjugate pair that would be best for preparing a buffer with a pH of 6.96 has to be given.

Concept Introduction:

Henderson-Hasselbalch equation:

The pH can be calculated using Henderson-Hasselbalch equation as follows,

pH = pKa + log[conjugate base][weak acid]pH = pKa + log[A-][HA]

Buffer:

A buffer is solution containing a mixture of two substances. A buffer solution is characterized by the tendency to resist changes in pH when some limited of amount of acid or base are added to it.

A buffer should possess either a weak acid and its conjugate anion that is a conjugate base (weak) or a weak base and its conjugate cation that is a conjugate acid (weak).

Explanation

To Give: The conjugate pair that would be best for preparing a buffer with a pH of 6.96

Given data:

NH4+(aq) + H2 H3O+(aq) + NH3(aq); Ka=5.6×1010H2CO3(aq) + H2 H3O+(aq)+ HCO3(aq); Ka=4

(b)

Interpretation Introduction

Interpretation:

From the given equilibrium equations and Ka value for the given three reaction systems,

The preparation of 100 mL of a buffer with a pH of 6.96 by assuming to have a 0.10 M solutions of each pair has to be explained

Concept Introduction:

Henderson-Hasselbalch equation:

The pH can be calculated using Henderson-Hasselbalch equation as follows,

pH = pKa + log[conjugate base][weak acid]pH = pKa + log[A-][HA]

Buffer:

A buffer is solution containing a mixture of two substances. A buffer solution is characterized by the tendency to resist changes in pH when some limited of amount of acid or base are added to it.

A buffer should possess either a weak acid and its conjugate anion that is a conjugate base (weak) or a weak base and its conjugate cation that is a conjugate acid (weak).

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