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# pH and pOH

## Introduction

Before discussing pH we must understand the equilibrium behavior of water.

H2O + H2O H3O+(aq) + OH-(aq)
or
H2O H+(aq) + OH-(aq)

The equilibrium constant expression for this reaction is:

Kw = [H3O+][OH-] = [H+][OH-] = 1.001x10-14 (at 25 oC, Kw is temperature dependent)

(Using [H3O+] is equivalent to using [H+].)

The equilibrium constant, Kw, is called the dissociation constant or ionization constant of water.

In pure water [H+] = [OH-] = 1.00x10-7 M.

## pH and pOH

Working with numbers like 1.00x10-7 M to describe a neutral solution is a rather inconvient. Acidity and basicity is described on a more convenient logarithmic scale:

pH is a shorthand notation for -log[H+]
and
pOH is a shorthand notation for -log[OH-].

Because of the equilibrium relationship between [H+] and [OH-], pH + pOH = 14.

Solutions are called
neutral when pH = 7, [H+] = [OH-] = 1.00x10-7
acidic when pH < 7, [H+] > 1.00x10-7
basic when pH > 7, [H+] < 1.00x10-7

Example: What is the pH of a solution of 0.025 M HNO3?

HNO3 is a strong acid and for all practical purposes dissociates completely.

HNO3(aq) + H2O NO3-(aq) + H3O+(aq)

[H+] = 0.025 M

pH = -log(0.025 M) = 1.6

What is the pOH of this solution? There are 2 ways to calculate pOH:

Kw = 1.00x10-14 = [0.025 M][OH-]
[OH-] = 4.00x10-13
pOH = -log(4.00x10-13) = 12.40

or:

pOH = 14.00 - pH = 14.0 - 1.60 = 12.40

What are pH and pOH for a 0.0025 M solution of HNO3?

pH = -log(0.0025 M) = 2.60
pOH = 14.00 - 2.60 = 11.40

Notice that pH and pOH change by 1 for a factor of 10 change in [H+] and [OH-].

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