5.5.3 The Ionic Product of Water

• In all aqueous solutions, an equilibrium exists in water where a few water molecules dissociate into protons and hydroxide ions
• We can derive an equilibrium constant for the reaction:

• This is a specific equilibrium constant called the ionic product for water
• The product of the two ion concentrations is always 1 x 10^-14 mol² dm^-6
• This makes it straightforward to see the relationship between the two concentrations and the nature of the solution:

[H⁺] & [OH^–] Table

Strong bases

• Strong bases 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
• 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