Chapter 16, Problem 16.37QP

### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343

Chapter
Section

### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343
Textbook Problem

# Boric acid, B(OH)3, is used as a mild antiseptic. What is the pH of a 0.015 M aqueous solution of boric acid? What is the degree of ionization of boric acid in this solution? The hydronium ion arises principally from the reaction B ( OH ) 3 ( a q ) + 2 H 2 O ( l ) ⇌ B ( OH ) 4 − ( a q ) + H 3 O + ( a q )

Interpretation Introduction

Interpretation:

The pH of Boric acid and its degree of ionization has to be calculated

Concept Information:

Acid ionization constant Ka :

The ionization of a weak acid HA can be given as follows,

HA(aq)H+(aq)+A-(aq)

The equilibrium expression for the above reaction is given below.

Ka=[H+][A-][HA]

Where Ka is acid ionization constant,

[H+]   is concentration of hydrogen ion

[A-]   is concentration of acid anion

[HA] is concentration of the acid

pH definition:

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

pH=-log[H+]

Degree of ionization is the amount of reactant that undergone ionization in the reaction.

To Calculate: The pH of Boric acid and its degree of ionization

Explanation

Given data:

Boric acid (B(OH)3) is used as a mild antiseptic.

The concentration of Boric acid = 0.015 M

Calculation of pH:

Explanation:

Set up the equilibrium table for the given acid solution.

Let x be the unknown concentration.

Represent, Boric acid as HBo and its conjugate anion as Bo

 HBo(aq)+H2O(l)     ⇄        H3O+(aq)  +   Bo-(aq) Initial (M) 0.015 -x 0.015-x 0.00 0.00 Change (M) +x +x Equilibrium (M) x x

The Ka of the Boric acid is calculated as follows,

Ka =[H3O+][Bo-][HBo] =(x)20

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