5.3.3 pH Calculations

Strong acids

• Strong acids are completely ionised in solution
  • This can also be described as dissociating in the ions

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

Strong bases

• Strong bases are completely ionised in solution
  • This can also be described as dissociating in the ions

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
• 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, simply by dividing K_w by the [H⁺]

Answer

Sodium hydroxide is a strong base which ionises as follows:

NaOH (aq) → Na⁺ (aq) + OH^- (aq) 

Answer 1:

   The pH of the solution is:

   [H⁺] = K_w  ÷ [OH^-]

   [H⁺] = (1 x 10^-14) ÷ 0.15 = 6.66 x 10^-14

   pH = -log[H⁺]

         = -log 6.66 x 10^-14  = 13.17

Answer 2

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

   pH = -log[H⁺]

   [H⁺]= 10^-pH

   [H⁺]= 10^-10.50

   [H⁺]= 3.16 x 10^-11 mol dm^-3

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

   K_w = [H⁺] [OH^-]

    [OH^-]= K_w  ÷  [H⁺] 

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

   Since K_w is 1 x 10^-14 mol² dm^-6,

    [OH^-]= (1 x 10^-14)  ÷  (3.16 x 10^-11)

           [OH^-]= 3.16 x 10^-4 mol dm^-3

Answer

The correct option is D

   Since K_w = [H⁺] [OH^–], rearranging gives [H⁺]  = K_w ÷ [OH^–]

   The concentration of  [H⁺] is (1 × 10⁻¹⁴) ÷ (1.0 × 10⁻³) = 1.0 × 10⁻¹¹ mol dm⁻³

   [H⁺]= 10^-pH

   So the pH = 11.00

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 = -log[H⁺]

   = -log(1.32 x 10^-3)

   = 2.88

We must make assumptions when calculating the pH of a weak acid

• [H⁺] at equilibrium is equal to the [A^-] at equilibrium because they have dissociated according to a 1:1 ratio
  
  • This is because the amount of H⁺ from the dissociation of water is insignificant
• The amount of dissociation is so small that we assume that the initial concentration of the undissociated acid has remained constant 
  • So initial [HA] is equal to the [HA] at equilibrium

The strength of acid

• The stronger the acid the greater the concentration of hydrogen ions in the solution at equilibrium
  • This corresponds to a larger value for K_a
  • If the acid is stronger, the dissociation will be greater, therefore the difference in the values for initial [HA] and [HA] at equilibrium will be greater