Acid Strength (Edexcel A Level Chemistry): Revision Note

Acid Dissociation

Strong acids

• A strong acid is an acid that dissociates almost completely in aqueous solutions

  • HCl (hydrochloric acid), HNO₃ (nitric acid) and H₂SO₄ (sulfuric acid)

• The position of the equilibrium is so far over to the right that you can represent the reaction as an irreversible reaction

The diagram shows the complete dissociation of a strong acid in aqueous solution

Weak acids

• A weak acid is an acid that partially (or incompletely) dissociates in aqueous solutions

  • Eg. most organic acids (ethanoic acid), HCN (hydrocyanic acid), H₂S (hydrogen sulfide) and H₂CO₃ (carbonic acid)

• The position of the equilibrium is more over to the left and an equilibrium is established

The diagram shows the partial dissociation of a weak acid in aqueous solution

Enthalpy change of neutralisation 

• The enthlapy change of neutralisation of strong acids and strong bases are very similar, around -57 to -58 kJmol^-1

• This is because the acids and alkalis are fully ionised and the neutralisation reaction between H⁺ + OH^– occurs to produce water:

  • H⁺ (aq) + OH^– (aq) → H₂O (l)

• The other ions are not involved in the reaction, i.e. are spectator ions, so do not affect neutralisation

• As this is the reaction that is occurring in each stong acid-strong alkali reaction, then the enthalpy change of neutralisation will be very similar

• Weak acids and weak alkalis only partially ionise, so energy has to be used to fully ionise them

• This means that the resulting enthalpy change of neutralisation will be less exothermic , i.e. less negative

  • For example, the standard enthalpy change of neturalisation of ethanoic acid with sodium hydroxide is -55.2 kJ mol^-1

Ka Expressions

• For weak acids as there is an equilibrium we can write an equilibrium constant expression for the reaction

• This constant is called the acid dissociation constant, K_a, and has the units mol dm^-3

• Values of K_a are very small, for example for ethanoic acid K_a = 1.74 x 10^-5 mol dm^-3 

• When writing the equilibrium expression for weak acids, the following assumptions are made:

  • The concentration of hydrogen ions due to the ionisation of water is negligible

• The value of K_a indicates the extent of dissociation

  • The higher the value of K_a the more dissociated the acid and the stronger it is

  • The lower the value of K_a the weaker the acid

pH Calculations for Acids

Strong acids

• Strong acids are completely ionised in solution

HA (aq) → H⁺ (aq) + A^- (aq)

• Therefore, the concentration of hydrogen ions, H⁺, is equal to the concentration of acid, HA

• The number of hydrogen ions formed from the ionisation of water is very small relative to the [H⁺] due to ionisation of the strong acid and can therefore be neglected

• The total [H⁺] is therefore the same as the [HA]

Answer

• [HCl] = [H⁺] = 0.01 mol dm^-3

  • pH = - log[H⁺]

  • pH = - log[0.01] = 2.00

The pH of dibasic acids

• Dibasic or diprotic acids have two replaceable protons and will react in a 1:2 ratio with bases

• Sulfuric acid is an example

   H₂SO₄ (aq)  + 2NaOH (aq) → Na₂SO₄ (aq) + 2H₂O (l)

• You might think that being a strong acid it is fully ionised so the concentration of the hydrogen is double the concentration of the acid

• This would mean that 0.1 mol dm^-3 would be 0.2 mol dm^-3 in [H⁺] and have a pH of 0.69

• However, measurements of the pH of  0.1 mol dm^-3 sulfuric acid show that it is actually about pH 0.98, which indicates it is not fully ionised

• The ionisation of sulfuric acid occurs in two steps

H₂SO₄ → HSO₄^- + H⁺

HSO₄^- ⇌ SO₄^2- + H⁺

• Although the first step is thought to be fully ionised, the second step is suppressed by the abundance of hydrogen ions from the first step creating an equilibrium

• The result is that the hydrogen ion concentration is less than double the acid concentration

Weak acids

The pH of weak acids can be calculated when the following is known: The concentration of the acid The K_a value of the acid From the K_a expression we can see that there are three variables:

• However, the equilibrium concentration of [H⁺] and [A^-] will be the same since one molecule of HA dissociates into one of each ion

• This means you can simplify and re-arrange the expression to

K_a x [HA] = [H⁺]²

[H⁺]² = K_a x [HA] 

• Taking the square roots of each side

[H⁺] = √(K_a x [HA])

• Then take the negative logs

pH = -log[H⁺] = -log√(K_a x [HA])

Answer

Ethanoic acid is a weak acid which ionises as follows:

CH₃COOH (aq) ⇌ H⁺ (aq) + CH₃COO^- (aq)

Step 1: Write down the equilibrium expression to find K_a

Step 2: Simplify the expression

The ratio of H⁺ to CH₃COO^- ions is 1:1

The concentration of H⁺ and CH₃COO^- ions are therefore the same

The expression can be simplified to:

Step 3: Rearrange the expression to find [H⁺]

Step 4: Substitute the values into the expression to find [H⁺]

= 1.32 x 10^-3 mol dm^-3

Step 5: Find the pH

pH = -log[H⁺]

= -log(1.32 x 10^-3)

= 2.88