   # A 0.040 M solution of an acid, HA, has a pH of 3.02 at 25°C. What is K a for this acid? (a) 23 × 10 −5 (b) 5.7 × 10 −4 (c) 2.4 × 10 −2 (d) 4.3 × 10 −10 ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 16.7, Problem 2RC
Textbook Problem
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## A 0.040 M solution of an acid, HA, has a pH of 3.02 at 25°C. What is Ka for this acid? (a) 23 × 10−5 (b) 5.7 × 10−4 (c) 2.4 × 10−2 (d) 4.3 × 10−10

Interpretation Introduction

Interpretation:

Value of acid-dissociation constant Ka for an orgainc acid is to be calculated.

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.)+ A1(aq.)

The dissociation constant for the acid is Ka ,

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

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

pKa=log(Ka)

pKa: It is introduced as an index to express the acidity of weak acids, where pKa is defined as fallows, we consider the Ka constant for acetic acid 1.8×105(pKa) but pKa constant for is 4.8 which is a simpler expression. Furthermore the smaller pKa value has strongest acid.

### Explanation of Solution

Given the statement acid dissociation constant of the (HA) is calculated as fallows

Given: The general equilibrium reaction takes place during acid dissociation is

HA (aq.) H+(aq.)+ A(aq.)

The ICE table is as follows,

EquilibriumHA(aq)H+(aq)+A(aq)Initial(M)0.04000Change(M)x+x+xEquilibrium(M)(0.04x)XX

The given acetic acid pH range has 3.02 at 250C

The pKa value of the acid is 895 . By using this Ka  is calculated as follows,

pKa=log(Ka)pKa=log(3.02)Hence[H+]=103.02(or)AntilogH+=10(pKa)=0.0009549Soxvalue9

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