The Ionic Product of Water (College Board AP® Chemistry)

The Ionic Product of Water

The autoionization constant of water, K_w

• The pH, pOH [H3O⁺] and [OH^-] are all related to one another

• An equilibrium exists in water, where a few water molecules dissociate into proton and hydroxide ions

2H₂O (l) ⇌ H₃O⁺ (aq) + OH^– (aq)

OR

H₂O (l) ⇌ H⁺ (aq) + OH^– (aq)

• The equilibrium expression for this is:

K_w = [H₃O⁺] [OH^–] = 1.00 x 10^-14

• In a neutral solution [H₃O⁺] = [OH^–]

• [H₃O⁺] can be calculated a 

  • At 25 °C the value of K_w is 1.00 x 10^-14

• Since the concentration of the H₃O⁺ and OH^- ions is very small, the concentration of water is considered to be a constant

• This means that the expression can be rewritten as:

K_w = [H₃O⁺] [OH^-]

• Where K_w =  1.00 10^-14 at 25 °C
  

• Taking the negative log for the equilibrium will give

pK_w = pH + pOH = 14

How does temperature affect K_w?

• The ionisation of water is an endothermic process

2H₂O (l) ⇌ H₃O⁺  (aq) + OH^- (aq) 

• In accordance with Le Châtelier's principle, an increase in temperature will result in the forward reaction being favoured

  • This causes an increase in the concentration of the hydrogen and hydroxide ions

  • This leads to the magnitude of K_w increasing

  • Therefore, the pH will decrease

• Increasing the temperature, decreases the pH of water (becomes more acidic)

• Decreasing the temperature, increases the pH of water (becomes more basic)

Graph to show how K_w changes with temperature

As temperature increases, K_w increases so pH decreases

Relationship between H⁺ (H₃O⁺) , OH^–, pH and pOH

To make a conversion, follow the arrow and equation given, so to convert OH^– (aq) to pOH use pOH = -log₁₀[OH^–]