   Chapter 17, Problem 87GQ

Chapter
Section
Textbook Problem

Describe the effect on the pH of the following actions or explain why there is not an effect: (a) Adding sodium acetate, NaCH3CO2, to 0.100 M CH3CO2H (b) Adding NaNO3 to 0.100 M HNO3

(a)

Interpretation Introduction

Interpretation:

pH of the solution has to be determined that contains 0.25moll1 acetic acid and 0.10moll1 sodium acetate.(Kaforaceticacid=1.8×105)

Concept introduction:

In aqueous solution an acid undergoes ionization. The ionization of an acid is 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.)

A weak acid undergo 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)

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 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”.

The pH of 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:

The value of dissociation constant for acetic acid is 1.8×105.

Initial concentration of acetic acid is 0.10moll1

The ICE table gives the relation between acetic and acetate ion concentration.

EquationCH3COOH(aq)+H2O(aq)H3O+(aq)+CH3COO(aq)Initial(M)0.1000Change(M)x+x+xEquilibrium(M)0.10xx+x

Approximation, the value of x is very small on comparison to 0.10, hence itcan be neglected with respect to 0.10.

The value of x i.e., the concentration of hydronium ion is be calculated by using equation  (1).

Ka=(x)(x)(c)

Substitute, 0.10moll1 for c and 1.8×105 for c.

1.8×105=(x)(x)0.10

Rearrange above relation for x

x=(1

(b)

Interpretation Introduction

Interpretation:

pH of the solution has to be determined that contains 0.25moll1 acetic acid and 0.10moll1 sodium acetate.(Kaforaceticacid=1.8×105)

Concept introduction:

In aqueous solution an acid undergoes ionization. The ionization of an acid is 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.)

A weak acid undergo 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)

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 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”.

The pH of solution is calculated by using the relation,

pH=log[H3O+] (3)

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