The Ionic Product of Water (DP IB Chemistry: SL)

pH of water

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

H₂O(l) ⇌ H⁺(aq) + OH^-(aq)

• The equilibrium constant for this reaction is:

K_c x [H₂O] = [H⁺] [OH^-]

• Since the concentration the H⁺ and OH^- ions is very small, the concentration of water is considered to be a constant, such that the expression can be rewritten as:

K_w = [H⁺] [OH^-]

Where K_w (ionic product of water)      =         K_c x [H₂O]

 =         10^-14 mol² dm^-6 at 298K

• The product of the two ion concentrations is always 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

Answer:

The correct option is D.

• • Since K_w = [H⁺] [OH^-] , rearranging gives [H⁺]  = K_w ÷ [OH^-]
  • The concentration of  [H⁺] is (1.0 × 10⁻¹⁴) ÷ (1.0 × 10⁻³) = 1.0 × 10⁻¹¹ mol dm⁻³
  • So the pH = 11