Chapter 14, Problem 54E

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

Chapter
Section

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

# The pOH of a sample of baking soda dissolved in water is 5.74 at 25°C. Calculate the pH, [H+], and [OH–] for this sample. Is the solution acidic or basic?

Interpretation Introduction

Interpretation:

• The pOH,[H+] and [OH] of the given sample of baking soda is to be calculated.
• The classification of this sample as acidic or basic is to be identified.

Concept introduction:

The pH of a solution is define as a figure that expresses the acidity of the alkalinity of a given solution.  A logarithmic scale is used on which, the value 7 corresponds to a neutral species, a value less than 7 corresponds to an acid and a value greater than 7 corresponds to a base.

The pH of a solution is calculated by the formula, pH=log[H+]

The sum, pH+pOH=14

The equilibrium constant for water is denoted by Kw and is expressed as,

Kw=[H+][OH]

Explanation

To determine: The pOH,[H+] and [OH] of the given sample of gastric juice and the classification of this sample as acidic or basic.

The [OH] is 1.8×10-6M_ .

Given

pOH=5.74

The pOH of a solution is calculated by the formula,

pOH=log[OH]

Rearrange the above expression to calculate the value of [OH] .

[OH]=10pOH

Substitute the given value of pH in the above expression.

[OH]=105.74=1.8×10-6M_

The [H+] is 5.5×10-9M_ .  The solution is basic.

The calculated value of [OH]=1.8×10-6M

The temperature is 25°C .

The [H+] is calculated by the formula,

Kw=[H+][OH]=1×1014

Substitute the given value of [OH] in the above expression

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