   Chapter 16, Problem 115IL

Chapter
Section
Textbook Problem

Nicotinic acid, C6H5NO2, is found in minute amounts in all living cells, hut appreciable amounts occur in liver, yeast, milk, adrenal glands, white meat, and corn. Whole wheat (lour contains about 60. 0μg per gram of flour. One gram (1.00 g) of the acid dissolves in water to give 60. mL of solution having a pH of 2.70. What is the approximate value of Ka for the acid? Nicotinic acid

Interpretation Introduction

Interpretation:

To calculate the Ka value for the Nicotinic acid.

Concept introduction:

The pH of a solution is measure of hydronium ion concentration in a solution.

pH=log[H3O+]                                                                                                     (1)

The reaction of the Nicotinic acid with water is an example of acid hydrolysis where water acts as a base which abstract proton from the carboxyl part of the Nicotinic acid.

Explanation

An equilibrium constant (K) is the ratio of the concentration of products and reactants raised to appropriate stoichiometric coefficient at equilibrium.

The dissociation constant of Nicotinic acid with water is written as follows,

C6H5NO2(aq)+H2O(l)C6H4NO2(aq)+H3O+(aq)

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

Ka=[C6H4NO2][H3O+][C6H5NO2]                                                                                       (2)

An equilibrium constant (K) with subscript a indicates that it is an equilibrium constant of an acid in water.

Given:

The pH of the solution is 2.70.

Mass of Nicotinic acid is 1g.

The Volume of the solution is 0.06L.

The Molecular weight of the Nicotinic acid is 123.11gmol1.

The number of moles (n) is equal to the mass of Nicotinic acid divided by its molecular weight.

n=1g123.11gmol1=0.008mol

The initial concentration of Nicotinic acid (M) is equal to the number of moles (n) divided by the volume of solution.

M=0.008mol0.06L=0.13molL1

From equation (1), calculating the concentration of hydronium ion,

2.70=log[H3O+][H3O+]=0

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