Ionic Product of Water (Edexcel IAL Chemistry)

Ionic Product of Water, Kw

Spec point — 14.10

• 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^-14mol² dm-6
• This makes it straightforward to see the relationship between the two concentrations and the nature of the solution:

[H⁺] & [OH^–] Table

[H+] and [OH-] table, downloadable IB Chemistry revision notes

The relationship between K_w and_* pK_w *is given by the following equation:

pK_w= -logK_w

pH Calculation of a Strong Base

Spec point — 14.11

• 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⁺]

Worked Example pH calculations of a strong alkali

Question 1:

Calculate the pH of 0.15 mol dm^-3 sodium hydroxide, NaOH

Question 2:

Calculate the hydroxide concentration of a solution of sodium hydroxide when the pH is 10.50

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