Chapter 17, Problem 4PS

### 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

# What is the pH of 0.15 M acetic acid to which 1.56 g of sodium acetate, NaCH3CO2 has been added?

Interpretation Introduction

Interpretation:

pH of 0.15M acetic acid solution to which 1.56 g sodium acetate is added has to be determined.

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) (1)

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 (1) 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

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) in equation (1).

The acid-dissociation constant will be

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

Ka=x2(cx) (2)

This table can be modified if one of the ion is already present before the ionization of acid. Then there will be some extent of suppression of dissociation of the weak 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 weak acid in presence of strong electrolyte having common ion is called as “Common Ion effect”.

Therefore a modified ICE table (2) is used to give the concentration relationships between ions. For example, if anion A 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

Here,

• y is the initial concentration of the anion A (common ion coming from strong electrolyte) present in the solution before the dissociation of a weak acid HA.

From the ICE table (2),

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

Substitute x for [H3O+](eq), x+y for [A](eq) and cx for [HA](eq) in equation (1).

The expression for  acid dissociation constant, Ka, will be given as,

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 x can be neglected with respect to y and c.

Then Ka can be written as,

Ka=(x)(y)c (3)

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

pH=log[H3O+] (4)

Explanation

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

Given:

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

The value of dissociation constant, Ka, for acetic acid is 1.8Ã—10âˆ’5.

The initial concentration of acetic acid is 0.15â€‰M.

The given mass of sodium acetate is 1.56Â g.

Acetic acid undergoes dissociation in aqueous solution and the reaction is given as,

Â Â Â Â CH3COOH(aq)+H2O(aq)â‡ŒH3O+(aq)+CH3COOâˆ’(aq)

Sodium acetate is a strong electrolyte and dissociates to give acetate ion and sodium ion in aqueous solution.

Â Â Â Â NaCH3CO2(aq)+H2O(l)â‡ŒCH3COOâˆ’(aq)+Na+(aq)

In presence of sodium acetate having common acetate ion, due to the common ion effect suppression of dissociation of acetic acid will occur.

Sodium acetate is a strong electrolyte, therefore, the concentration of acetate ion coming from sodium acetate is equal to the initial concentration of the sodium acetate and calculated as follows,

[NaCH3CO2]=w(M)(V)Â  (5)

Here,

• w is the given mass of sodium acetate.
• M is the gram molecular mass of the sodium acetate.
• VÂ  is the volume of the solution in liter.

Gram molecular mass of sodium acetate is 82.034Â gâ‹…molâˆ’1.

Substitute 1.56â€‰g for w, 82.034Â gâ‹…molâˆ’1 for M and 1Â L for VÂ  in equation (5).

[NaCH3CO2]=1.56â€‰g(82.034Â gâ‹…molâˆ’1)(1Â L)Â =0.019Â molâ‹…Lâˆ’1=0.019Â M

The ICE table (3) is given as follows,

EquationCH3COOH(aq)+H2O(aq)â‡ŒH3O+(aq)+CH3COOâˆ’(aq)Initial(M)0

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