# Chapter 16, Problem 16.86QP

FindFindarrow_forward

### CHEMISTRY: ATOMS FIRST VOL 1 W/CON...

14th Edition
Burdge
ISBN: 9781259327933

#### Solutions

Chapter
Section
FindFindarrow_forward

### CHEMISTRY: ATOMS FIRST VOL 1 W/CON...

14th Edition
Burdge
ISBN: 9781259327933
Interpretation Introduction

Interpretation:

The pH of a 0.25 M aqueous solution H3PO4 has to be calculated.

Concept Information:

Acid ionization constant Ka :

Acids ionize in water. Strong acids ionize completely whereas weak acids ionize to some limited extent.

The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization.

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 concentration of hydrogen ion is measured using pH scale.  The acidity of aqueous solution is expressed by pH scale.

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

pH=-log[H3O+]

Diprotic and polyprotic acids:

Acids having two or more hydrogen atoms are termed as diprotic or polyprotic acids.  These acids lose one proton at a time by undergoing successive ionizations.

For diprotic acids, the successive ionization constants are designated as Ka1andKa2

For triprotic acids, the successive ionization constants are designated as Ka1,Ka2andKa3

To Calculate: The pH of a 0.25 M aqueous solution H3PO4

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Answers to Your Study Problems

Solve them all with bartleby. Boost your grades with guidance from subject experts covering thousands of textbooks. All for just \$9.99/month

Get As ASAP