   Chapter 13, Problem 72QAP ### Chemistry: Principles and Reactions

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

#### Solutions

Chapter
Section ### 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 2.3×108.

The ionization reaction is shown as follows:

NaHCO3(aq)+H2OH2CO3(aq)+NaOH(aq)

The ICE table for above reaction will be as follows:

NaHCO3(aq)+H2OH2CO3(aq)+NaOH(aq)I           0.72                     -                     -              -C          -x                        -                     x            xE         0.72-x                                        x            x

The expression for base dissociation constant will be as follows:

Kb=[H2CO3][NaOH][NaHCO3]

Putting the values,

2.3×108=(x)(x)(0.72x)

On rearranging,

x2+2.3×108x1.656×108=0

Comparing it with quadratic equation as follows;

ax2+bx+c=0

The value of x can be calculated as follows:

x=b±b24ac2a

Putting the values,

x=(2

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