8.3 The pH Scale

Background knowledge.

It is assumed that you have a working knowledge of the pH scale from GCSE, or equivalent.

So you should know the significance of the numbers and an approximate colour for each pH using Universal Indicator.

You should also know a little about other indicators.

It’s not necessary to know the pH range over which these indicators change but you should know their colours in strong acid and strong alkali.

Acidity (or alkalinity) of a solution is due to the concentration of H⁺ ions, [H⁺].

But [H⁺] is always a very small number.

Eg. For a solution of Sodium Hydroxide [H⁺] may be 0.00000000000001

This is best expressed in Standard Form as [H⁺] = 1 x 10^-14.

But no one likes Standard Form so we take Log₁₀ [H⁺] to get a simpler number

Log₁₀ [1 x 10^-14] = -14

But negative numbers are a pain so we define pH as

pH = -Log₁₀[H⁺]

This is all you have to write as a definition – no explanation is necessary.

What are the pHs of solutions with the following concentrations of H⁺ ions?

I. [H⁺] = 1 x 10^-4

II. [H⁺] = 1.8 x 10^-4

III. [H⁺] = 1.8 x 10^-12

IV. [HCl] = 9.8 x 10^-6

V. [H₂SO₄] = 9.8 x 10^-6

VI. [HNO₃] = 7.78 x 10^-2

VII. [H₂SO₄] = 7.78 x 10^-7

The opposite function to Log is Antilog (or Inverse Log).

You’ll find it as the 2^nd Function of the Log button, but it will probably say 10^x.

Since pH = -Log₁₀[H⁺]

[H⁺] = 10^-pH

So, a solution with pH = 8 has [H⁺] = 10^-8 (or 1x 10^-8)

1 What are the [H⁺]’s of solution with the following concentrations of pH?

I. pH = 9.42

II. pH = 7.12

III. pH = 3.4

2. Find....

IV. [HCl] with pH = 12.2

V. [H₂SO₄] with pH = 9.12

VI. [H₃PO₄] with pH = 8.92

Note:

The power of 10 in [H⁺] is always close to the -pH.

If it isn’t, you’ve pressed the wrong buttons!

As we’ve seen, finding the pH of a strong acid is easy because [H⁺] is equal to [HCl] & [HNO₃], and equal to ½ [H₂SO₄].

But, for a strong base like Sodium Hydroxide we only know [OH^-] = [NaOH].

Fortunately, autoionisation of water gives a constant K_c, like other Equilibria

K_c = [H⁺][OH^-] / [H₂O]

So little autoionisation happens that [H₂O] is constant at every temp, so we write

K_w = [H⁺][OH^-]

You should know that at 25^oC K_w = 1 x 10^-14- -- So [H⁺] = 1 x 10^-7

And so neutral pH at standard temperature is 7.

1 What is the pH of a solution of 2.5 x 10^-2 moldm^-3 NaOH

[OH^-] = [NaOH] = 2.5 x 10^-2 moldm^-3

K_w = [H⁺][OH^-]

1 x 10 ^-14 = [H⁺] x 2.5 x 10^-2 [H⁺] = 4 x 10 ^-13

pH = Log [4 x 10 ^-13] = 12.4

2 What is the pH of a solution of 2.5 x 10^-2 moldm^-3 Ca(OH)₂

[OH^-] =2 x [Ca(OH)₂] = 5 x 10^-2 moldm^-3

K_w = [H⁺][OH^-]

1 x 10 ^-14 = [H⁺] x 5 x 10^-2 [H⁺] = 2 x 10 ^-13

pH = Log [2 x 10 ^-13] = 12.7

1. Find the pH of the following solutions

a. 1.2 mol/dm³ NaOH

b. 1.2 x 10 ^-3 mol/dm³ NaOH

c. 1.2 mol/dm³ Ca(OH)₂

d. 5.4 x 10 ^-2 Ca(OH)₂

2. Find [OH^-] in solutions of:

a. 3.2 mol/dm³ HCl

b. 3.2 x 10 ^-2 mol/dm³ HNO₃

c. 0.2 mol/dm³ H₂SO₄

d. 7.6 x 10 ^-2 H₂SO₄