   Chapter 16, Problem 98GQ

Chapter
Section
Textbook Problem

Chloroacetic acid, ClCH 2 CO 2 H , is a moderately weak acid ( K a = 1 .40×10 − 3 ) . If you dissolve 94.5 mg of the acid in water to give 125 ml of solution, what is the pH of the solution?

Interpretation Introduction

Interpretation:

The pH of the chloroacetic acid solution has to be determined.

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 an 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]

Ion product constant for water

Kw= [H3O+][OH-]       =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]

Relation between pH and pOH

pH + pOH =14

Explanation

Let’s calculate the molarity of chloroacetic acid.

The given mass of chloroaceticacid = 94.5mgconvert mg into grams                    = 94.5mg × 1g1000mg                    = 0.0945g

And,

The given volume of solution = 125mL                    =125ml×1L1000mL                    = 0.125L

Concentration of chloroceticcid =moles of chloroaceticacidVolume of solution in Litres                     =[0.0945g ClCH2CO2H94.5g/mol ClCH2CO2H]0.125L                     =0.008M

The equilibrium chemical reaction of chloroacetic acid is as follows.

ClCH2CO2H(aq) + H2O(l)ClCH2CO2-(aq) + H3O+(aq)

Equilibrium expression:

Ka[ClCH2CO2-][H3O+][ClCH2CO2H]

The initial concentration =0.008

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.

ClCH2CO2H(aq) + H2O(l)ClCH2CO2-(aq) + H3O+(aq)I         0.008M                     --                    --                    --C           -x                           --                  + x                 + xE        (0

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