   # Consider 0.10 M solutions of the following compound: AlCl 3 , NaCN, KOH, CsClO 4 , and NaF. Place these solutions in order of increasing pH. ### 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 169CWP
Textbook Problem
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## Consider 0.10 M solutions of the following compound: AlCl3, NaCN, KOH, CsClO4, and NaF. Place these solutions in order of increasing pH.

Interpretation Introduction

Interpretation: The 0.10M solutions of the given compounds are to be arranged in increasing order of pH.

Concept introduction: Strong bases dissociate completely and rapidly when dissolved in an aqueous solution. They have a very large value of Kb and a lower value of pH. The weak bases dissociate partially and slowly in an aqueous solution and have a very low value of Kb.and a higher value of pH.

To determine: The correct order of the 0.10M solutions of the given compounds in increasing order of pH.

### Explanation of Solution

Explanation

The compound NaCN 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.

NaCN+H2ONaOH+HCNInitial(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=[NaOH][HCN][NaCN]

Substitute the value of [NaCN], [NaOH] and [HCN] from ICE table in the above expression.

2.1×105=x×x0.10xx2+2.1×1052.1×106=0x=4.57×103

Thus, the value of [OH] is 4.57×103M.

The value of pOH is calculated by the formula.

pOH=log[OH]

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

pOH=log[4.57×103]pOH=2.34

The sum,

pH+pOH=14

Substitute the calculated pOH value in the above expression.

pH+2.34=14pH=142.34pH=11.66

The strong salts like Al(Cl)3 are formed by the reaction between a strong acid and a strong base. Al(Cl)3 is made up by the combination of a strong acid, that is HCl and a strong base, that is Al(OH)3. Hence, the salt is completely neutral.

The compound NaF 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.

NaF+H2ONaOH+HFInitial(M):0

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