Chapter 17.1, Problem 17.1CYU

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

You have a 0.30 M solution of formic acid (HCO2H) and have added enough sodium formate (NaHCO2) to make the solution 0.10 M in the salt. Calculate the pH of the formic acid solution before and after adding solid sodium formate.

Interpretation Introduction

Interpretation:

For the given solution 0.30molL1 formic acid, the value of pH has to be calculated before and after the addition of 0.10moll1 sodium formate solution.

Concept introduction:

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

HA(aq.)+ H2O(l)H3O+(aq.)+ A1(aq.)

A weak acid undergoes partial dissociation, there is a relation between dissociation constant Ka (or equilibrium constant) and the concentration of reactants and products.

Ka=[H3O+](eq)[A](eq)[HA](eq)

Here,

• [H3O+](eq) is the equilibrium concentration of hydronium ion.
• [A](eq) is the equilibrium concentration of conjugate base of the acid.
• [HA](eq) is the equilibrium concentration of acid.

The ICE table gives the relationship between the concentrations of species at equilibrium.

EquationHA(aq)+H2O(aq)H3O+(aq)+AInitial(M)c00Change(M)x+x+xEquilibrium(M)cxxx

The acid-dissociation constant will be

Ka=[H3O+](eq)[A](eq)[HA](eq)

From the ICE table (1),

[H3O+](eq)=x[A](eq)=x[HA](eq)=cx

Substitute x for [H3O+](eq),  x for [A](eq) and cx for [HA](eq).

The acid-dissociation constant will be

Ka=(x)(x)(cx)=x2(cx)

Ka=x2(cx) (1)

This relation can be modified if one of the ion is already present before the ionization of acid. Then there will be some extent of suppression for the acid. This can be explained on the basis of Le-Chatelier’s principle. According to which reaction will be more on the left side rather than right if one the ion from product side is already present before equilibrium. This suppression of ionization of acid is called as “Common Ion effect”.

Therefore an ICE table is used to give the concentration relationships between ions. For example, if anion A(y M) is already present in the solution before equilibrium,

EquationHA(aq)+H2O(aq)H3O+(aq)+A(aq)Initial(M)c0yChange(M)x+x+xEquilibrium(M)cxxy+x

The acid-dissociation, Ka, constant for such cases will be written as

Ka=[H3O+](eq)[A](eq)[HA](eq)

From ICE table

[H3O+](eq)=x[A](eq)=x+y[HA](eq)=cx

Substitute for Ka, x for [H3O+](eq),(x+y) for [A](eq) and (cx) for [HA](eq) in equation (3).

The acid-dissociation constant will be

Ka=(x)(x+y)(cx)

There is an assumption for common ion effect, according to which the value of “x” is very small on comparing to the initial concentration of acid (HA)c”and initial concentration of anion (A),“y”. Thus it can be neglected and the Ka can be written as

Ka=(x)(y)(cx)

The final relation for Ka in presence of a common ion, and neglecting x with respect to c is written as follows,

Ka=(x)(y)(c) (2

The pH of the solution is calculated by using the relation,

pH=log[H3O+] (3)

Explanation

The value of pH for the given solution is calculated as below.

Given:

Refer to table 16.2 in the textbook for the value of Ka

The value of dissociation constant Ka, for formic acid is 1.8Ã—10âˆ’4.

The initial concentration of formic acid is 0.30â€‰molâ‹…Lâˆ’1

The initial concentration of sodium formate is 0.10â€‰molâ‹…Lâˆ’1

The ICE table (1), for the solution when only formic acid is present and it undergoes dissociation.

EquationHCOOH(aq)+H2O(aq)â‡ŒH3O+(aq)+HCOOâˆ’(aq)Initial(M)0.3000Change(M)âˆ’x+x+xEquilibrium(M)0.30âˆ’xxx

From the ICE table (1),

(câˆ’x)=(0.30âˆ’x)

The acid dissociation constant for the formic acid will be written as,

Ka=x2(câˆ’x)

Substitute, (0.30âˆ’x) for (câˆ’x) and 1.8Ã—10âˆ’4 for Ka.

1.8Ã—10âˆ’4=x2(0.30âˆ’x)

Apply the approximation, that the value of x is very small in comparison to 0.30 so it can be neglected with respect to 0.30.

1.8Ã—10âˆ’4=x2(0.30)

Rearrange for x

x=(1.8Ã—10âˆ’4)(0.30)=(0.54)(10âˆ’2)=0.73Ã—10âˆ’2

Therefore the hydronium ion concentration [H3O+] is 0.73Ã—10âˆ’2â€‰molâ‹…Lâˆ’1.

Now using the equation (3) the value of pH is calculated as follows,

pH=âˆ’log[H3O+]

Substitute,0.73Ã—10âˆ’2â€‰ for [H3O+].

pH=âˆ’log(0.73Ã—10âˆ’2â€‰)=âˆ’(âˆ’2.13)=2

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started