pH & [H+] Calculations (CIE A Level Chemistry)

• If the concentration of H⁺ of an acid or alkali is known, the pH can be calculated using the equation:

pH = -log [H⁺]

• Similarly, the concentration of H⁺ of a solution can be calculated if the pH is known by rearranging the above equation to:

[H⁺] = 10^-pH

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 ([H⁺]) 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]

Worked Example: pH calculations of a strong acid

Answer

Hydrochloric acid is a strong monobasic acid

HCl (aq) → H⁺ (aq) + Cl^- (aq)

Answer 1

pH = -log [H⁺]

= -log 1.6 x 10^-4

= 3.80

Answer 2

The hydrogen concentration can be calculated by rearranging the equation for pH

pH = -log [H⁺]

[H⁺] = 10^-pH

= 10^-3.1

= 7.9 x 10^-4 mol dm^-3

Strong alkalis

• Strong alkalis are completely ionised in solution

BOH (aq) → B⁺ (aq) + OH^- (aq)

• Therefore, the concentration of hydroxide ions ([OH^-]) is equal to the concentration of base ([BOH])
  • Even strong alkalis have small amounts of H⁺ in solution which is due to the ionisation of water

• The concentration of OH^- in solution can be used to calculate the pH using the ionic product of water

K_w = [H⁺] [OH^-]

• Since K_w is 1.00 x 10^-14 mol² dm^-6

• Once the [H⁺] has been determined, the pH of the strong alkali can be founding using pH = -log[H⁺]
• Similarly, the ionic product of water can be used to find the concentration of OH^- ions in solution if [H⁺] is known

Worked Example: pH calculations of a strong alkali

Answer

Sodium hydroxide is a strong base which ionises as follows:

NaOH (aq) → Na⁺ (aq) + OH^- (aq) 

Answer 1

pH = -log [H⁺]

= -log 3.5 x 10^-11

= 10.5

Answer 2

•  Step 1: Calculate hydrogen concentration by rearranging the equation for pH

pH = -log [H⁺]

= 10^-pH

= 10^-12.3

= 5.01 x 10^-13 mol dm^-3

• Step 2: Rearrange the ionic product of water  to find the concentration of hydroxide ions

K_w = [H⁺] [OH^-]

• Step 3: Substitute the values into the expression to find the concentration of hydroxide ions

Since K_w is 1.00 x 10^-14 mol² dm^-6

= 0.0199 mol dm^-3

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

Worked Example: pH calculations of weak acids

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 = -log₁₀ [H⁺]

= -log₁₀ 1.32 x 10^-3

= 2.88