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Chapter 13, Problem 72QAP
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Chemistry: Principles and Reactions

8th Edition
William L. Masterton + 1 other
ISBN: 9781305079373

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Chapter
Section
BuyFindarrow_forward

Chemistry: Principles and Reactions

8th Edition
William L. Masterton + 1 other
ISBN: 9781305079373
Textbook Problem

Determine the [OH-] and pH of a 0.72 M solution of NaHCO3.

Interpretation Introduction

Interpretation:

The concentration of hydroxide ion and pH of the 0.72 M NaHCO3 solution should be determined.

Concept introduction:

The dissociation reaction of a weak base is represented as follows:

BOHB++OH

The expression for the base dissociation constant will be as follows:

Ka=[B+][OH][BOH]

Here, [B+] is equilibrium concentration of conjugate acid, [OH] is equilibrium concentration of hydroxide ion and [BOH] is the equilibrium concentration of weak base.

From the hydroxide ion concentration, hydrogen ion concentration can be calculated as follows:

[H+]=Kw[OH]

From the hydrogen ion concentration, pH of the solution can be calculated as follows:

pH=log[H+]

Explanation

The molarity of sodium hydrogen carbonate is given 0.72 M. The base dissociation constant it is 2.3Ã—10âˆ’8.

The ionization reaction is shown as follows:

NaHCO3(aq)+H2Oâ‡ŒH2CO3(aq)+NaOH(aq)

The ICE table for above reaction will be as follows:

Â Â Â Â Â Â Â Â Â Â NaHCO3(aq)+H2Oâ‡ŒH2CO3(aq)+NaOH(aq)IÂ Â Â Â Â Â Â Â Â Â Â 0.72Â 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 Â Â Â Â Â Â Â Â Â Â -CÂ Â Â Â Â Â Â Â Â Â -xÂ 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 xÂ Â Â Â Â Â Â Â Â Â Â Â xEÂ Â Â Â Â Â Â Â Â 0.72-xÂ 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 xÂ Â Â Â Â Â Â Â Â Â Â Â x

The expression for base dissociation constant will be as follows:

Kb=[H2CO3][NaOH][NaHCO3]

Putting the values,

2.3Ã—10âˆ’8=(x)(x)(0.72âˆ’x)

On rearranging,

x2+2.3Ã—10âˆ’8xâˆ’1.656Ã—10âˆ’8=0

Comparing it with quadratic equation as follows;

ax2+bx+c=0

The value of x can be calculated as follows:

x=âˆ’bÂ±b2âˆ’4ac2a

Putting the values,

x=âˆ’(2

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