   Chapter 16, Problem 32PS

Chapter
Section
Textbook Problem

Which is the stronger of the following two acids? (a) acetic acid, CH3CO2H, Ka = 1.8 × 10−5 (b) chloroacetic acid, ClCH2CO2H, pKa = 2.85

Interpretation Introduction

Interpretation: The determination of acidic strength of the acids has to be done by using their lower pKa values.

Concept introduction: In aqueous solution an acid undergoes ionization. The ionization of an acid is can be expressed in terms of equilibrium constant. The quantitative measurement tells about the strength of the acid. Higher the value of Ka stronger will be the acid. The acid dissocition can be represented as following equilibrium,

HA(aq.)+ H2O(l)H3O+(aq.)+ A(aq.)

The dissociation constant for the acid is Ka,

Ka=[H3O+][A][HA] s

For simplifications pKa value is used to find the acidic strength of acid. Which is calculated by taking negative logarithm of Ka.

pKa=-log(Ka)

The lower value of pKa of an acid, stronger will be the acid. Thus on comparing the pKa value it can be determined which acid is stronger one among them.

Explanation

On comparing the pKa values of both acids it is found that 2-Chloroacetic acid (ClCH2CO2H) has lower value of pKa than Acetic acid(CH3CO2H). Thus, 2-Chloroacetic acid (ClCH2CO2H) is stronger acid than Acetic acid(CH3CO2H).

Given: The pKa value for 2-Chloroacetic acid (ClCH2CO2H) is 2.8, and Ka value for acetic acid is given 1.8×105

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