   # Rank the following 0.10 M solutions in order of increasing pH. a. NH 3 b. KOH c. HC 2 H 3 O 2 d. KCl e. HCl ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 13, Problem 166CWP
Textbook Problem
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## Rank the following 0.10 M solutions in order of increasing pH.a. NH3b. KOHc. HC2H3O2d. KCle. HCl

Interpretation Introduction

Interpretation: The given solutions are to be arranged in increasing order of pH.

Concept introduction: Strong acids dissociate completely and rapidly when dissolved in an aqueous solution and have a very large value of Ka and a lower value of pH while the weak acids dissociate partially and slowly in an aqueous solution and have a very low value of Ka.and a higher value of pH.

### Explanation of Solution

Explanation

The [OH] of given NH3 is 4.15×10-4M_.

The compound NH3 is a weak base and establishes equilibrium upon dissociation in an aqueous solution. The [OH] value is calculated by the formation of an ICE (Initial Concentration Equilibrium) table.

NH3+H2ONH4++OHInitial(M):0.1000Change(M):xxxEquilibrium(M):0.10xxx

The equilibrium constant expression is calculated by the formula,

Kb=[Concentrationofproducts][Concentrationofreactants]

The equilibrium constant expression for the stated reaction is,

Kb=[NH4+][OH][NH3]

The value of Kb is 1.8×105.

Substitute the value of Kb, [NH4+], [OH] and [NH3] from ICE table in the above  expression.

1.8×105=x×x0.10xx2+1.8×1051.8×107=0x=4.15×10-4M_

Thus, the value of [OH] is 4.15×10-4M_.

The pOH of given NH3 solution is 3.38_.

The pOH value is calculated by the formula,

pOH=log[OH]

Substitute the value of [OH] in the above formula.

pOH=log[4.15×104]pOH=3.38_

The pH of given NH3 is 10.62_.

The sum,

pH+pOH=14

Substitute the calculated value of pOH in the above equation.

pH+3.38=14pH=143.38pH=10.62_

The pH of KCl is 7_.

The strong salts like KCl are formed by the reaction between a strong acid and a strong base. KCl is made up by the combination of a strong acid, that is HCl and a strong base, that is KOH.

These are strong salts and dissociate completely in an aqueous. The strong salts like KCl are neutral in nature and have pH value equal to 7_.

The [OH] value of KCl is. 0.10M_.

The given compound KOH is a strong base and dissociates completely and readily in an aqueous solution. The dissociation reaction that takes place is,

KOHK++OH

Thus, the given concentration of KOH will correspond to the [OH]. Therefore, the value of [OH] is 0.10M_.

The pOH of given KCl solution is 1_.

The pOH value is calculated by the formula,

pOH=log[OH]

Substitute the value of [OH] in the above formula

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