   Chapter 14, Problem 149AE

Chapter
Section
Textbook Problem

# At 25°C, a saturated solution of benzoic acid (Ka = 6.4 × 10−5) has a pH of 2.80. Calculate the water solubility of benzoic acid in moles per liter.

Interpretation Introduction

Interpretation: The acid dissociation constant and pH of benzoic acid solution is given. By using these values, the water solubility of benzoic acid inmolesperliter is to be calculated.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of concentration, it is represented as K . pH is the measure of H+ ions.

Explanation

Explanation

To determine: The water solubility of benzoic acid.

The stated reaction is,

C6H5COOH(aq)+H2O(l)C6H5COO(aq)+H3O+(aq)

The concentration of H+ ions is 0.001585_ .

Given

The acid dissociation constant is 6.4×105 .

The pH value of solution is 2.80 .

Formula

The pH is calculated using the formula,

pH=log10[H+]

Where,

• pH measure of H+ ions.
• [H+] is concentration of H+ ions.

Substitute the value of pH in the above equation.

pH=log10[H+]2.80=log10[H+][H+]=0.001585_

The stated reaction is,

C6H5COOH(aq)+H2O(l)C6H5COO(aq)+H3O+(aq)

The water solubility of benzoic acid is 0.040787M_ .

It is assumed that the initial concentration of benzoic acid C6H5COOH(aq) is x .

Make the ICE table for the above reaction

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