L4 - pH Scale

objectives

Understandings:

• pH = -log₁₀[H⁺] and [H⁺] = 10^-pH.

• A change of one pH unit represents a 10-fold change in the hydrogen ion concentration [𝐻+].

• pH values distinguish between acidic, neutral and alkaline solutions.

• The ionic product constant, 𝐾𝑤 = [H+][OH−] = 1x10^-14 at 298 K.

Applications and skills:

• Solving problems involving pH,[H+] and [OH−].

• Students should be familiar with the use of a pH meter and universal indicator.

Guidance:

• Students will not be assessed on pOH values.

• Students should be concerned only with strong acids and bases in this sub- topic.

• Knowing the temperature dependence of 𝐾w is not required.

• Equations involving H3O+ instead of H+ may be applied.

[H+] Concentration

• Introduction hydrogen ion concentration determines the acidity of a solution
• hydroxide ion concentration determines the alkalinity
• for strong acids and bases the concentration of ions is very much
• larger than their weaker counterparts which only partially dissociate.

Defining pH

Calculating pH of Strong Acids

It is easy to calculate the pH; you only need to know the concentration.

Calculate the pH of 0.02M HCl

HCl completely dissociates in aqueous solution HCl → H⁺ + Cl¯

One H⁺ is produced for each HCl dissociating so

[H⁺] = 0.02M = 2 x 10^-2 mol dm^-3

pH = - log [H⁺] = 1.70

Calculate the pH of 0.02M H₂SO₄

H₂SO₄ completely dissociates in aqueous solution H₂SO₄ → 2H⁺ + SO₄^2-

Two H⁺ are produced for each mole of H₂SO₄ dissociating so:

[H⁺] = 0.02M x 2 = 4 x 10^-2 mol dm^-3

pH = - log [H⁺] = 1.40

Ionic Product of Water K_w

= 1 x 10^-14 mol² dm^-6 (at 25°C)

Water is a very weak electrolyte with a dissociation equation of:

H₂O ⇌ H⁺ + OH^-

Kc = [H⁺] [OH^-]

[H₂O]

As it only weakly disassociates [H₂O] is constant so can be removed from K_c.

Therefore the K_c for this reaction changes to become K_w

K_w = [H⁺(aq)] [OH¯(aq)]

K_w = [H⁺] [OH^-]

The Ionic Product of Water shows us how the concentration of [H+] or [OH-] can be derived from a few known quantities.

As the pH of water is 7 at 298K we can derive that [H+] = 1x10^-7

K_w = 1x10^-14

1x10^-14 = [H⁺] [OH^-]

Therefore [H⁺] = 1x10^-7

[H⁺] must equal [OH^-] therefore [OH^-] = 1x10^-7

As the pH of water is always 7 at 298K we can therefore infer that must always equal K_w = 1x10^-14

This means that if we know the [H⁺] concentration then we can work out the concentration of [OH^-] as the sum of the Exponentials must multiply to 1x10^-14

Therefore if we have [H⁺] concentration of 1x10^-5 then the [OH^-] concentration must be 1x10^-9 in order to complete K_w = 1x10^-14

pH of Strong Bases

Calculate the pH of 0.1M NaOH

NaOH completely dissociates in aqueous solution NaOH --> Na⁺ + OH¯

One OH¯ is produced for each NaOH dissociating [OH¯] = 0.1M = 1 x 10^-1 mol dm^-3

The ionic product of water (at 25°C) K_w = [H⁺][OH¯] = 1 x 10^-14 mol² dm^-6

therefore [H⁺] = K_w / [OH¯] = 1 x 10^-13 mol dm^-3

pH = - log [H⁺] = 13