   Chapter 16, Problem 64PS

Chapter
Section
Textbook Problem

The sodium salt of propionic acid, NaCH 3 CH 2 CO 2 is used as an antifungal agent by veterinarians. Calculate the equilibrium concentration of H 3 O + and OH‒ the pH for a solution of 0.10M NaCH 3 CH 2 CO 2 .

Interpretation Introduction

Interpretation:

The equilibrium concentration of H3O+ and OH- the pH for a solution of 0.10M NaCH3CH2CO2 has to be calculated.

Concept introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka is acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb is base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

Ionic product constant for water  Kw= [H3O+][OH-]       =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]Reltion between pH and pOH pH + pOH =14

Explanation

NaCH3CH2CO2 dissociates into sodium Na+ and CH3CH2COO-. The chemical equilibrium reaction is as follows.

NaCH3CH2CO2(aq) Na+(aq) + CH3CH2COO(aq)

Sodium ions don’t affect the pH of the solution. The pH of the solution is completely depends upon the concentration of CH3CH2COO- ions.

But the resultant solution is basic in nature.

The CH3CH2COO- ions react with water, the equilibrium chemical reaction is as follows.

CH3CH2COO-(aq) + H2O(l) CH3CH2CO2H(aq) + OH-

The equilibrium expression:

Kb[CH3CH2CO2H][OH-][CH3CH2CO2]

Kb of CH3CH2CO2- is 7.7×10-10

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.

CH3CH2COO-(aq) + H2O(l) CH3CH2CO2H(aq) + OH-(aq)   I                 0.10                     --                    --                              --C                 -x                        --                  +x                            +x    E             (0

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