K_w Ion Product for Water Chemistry Tutorial

Key Concepts

• Water undergoes self-dissociation, also referred to as autodissociation or self-ionization (self-ionisation).

  H₂O_(l) ⇋ H⁺_(aq) + OH^-_(aq)

• The mass-action expression for the self-dissociation of water is:

  Q =	[H+(aq)][OH-(aq)]
  	[H2O(l)]

  [H⁺_(aq)] = concentration of hydrogen ions in aqueous solution
  [OH^-_(aq)] = concentration of hydroxide ions in aqueous solution
  [H₂O_(l)] = concentration of liquid water

• The concentration of liquid water, [H₂O_(l)], is essentially constant at any given temperature, and is incorporated into the equilibrium constant:

  K_w = [H⁺_(aq)][OH^-_(aq)]

• K_w is called the ion product for water or the dissociation constant for water
  (or the ionisation constant, or ionization constant, for water).
• K_w, like all equilibrium constants, is temperature dependent.
• For any given temperature, [H⁺_(aq)] = [OH^-_(aq)] for water, so water is neutral at all temperatures.
• For any given temperature, pH = pOH for water.
  Please note:

  Water is only pH = 7 at 25°C (≈ 298K)
  Water is only pOH = 7 at 25°C (≈ 298K)
  pH + pOH = 14 is only true at 25°C (≈ 298K)

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K_w Calculations at 25^oC (298K)

Question 1: Calculate the hydroxide ion and hydrogen ion concentration of water at 25°C (~298K), K_w = 10^-14.

Solution:

1. Write the equation for the ion product for water:

   K_w = [H⁺][OH^-]

2. Substitute in the value for K_w:

   10^-14 = [H⁺][OH^-]

3. Since [H⁺] = [OH^-]

   10^-14 = [H⁺]²

4. Take the square root of both sides of the equation to find [H⁺]:

   √10^-14 = √[H⁺]²
   √10^-14 = [H⁺]
   [H⁺] = 10^-7 mol L^-1

5. Since [H⁺] = [OH^-],

   [OH^-] = 10^-7 mol L^-1

Question 2: Calculate the pH and pOH of water at 25°C (298K).

Solution:

1. Write the equation for calculating pH:

   pH = -log₁₀[H⁺]

2. Substitute in the value for [H⁺] calculated above:

   pH = -log₁₀[10^-7] = 7

3. Write the equation for calculating pOH:

   pOH = -log₁₀[OH^-]

4. Substitute in the value for [OH^-] calculated above:

   pOH = -log₁₀[10^-7] = 7

Temperature Dependence of K_w

The reaction

H₂O_(l) ⇋ H⁺_(aq) + OH^-_(aq)

is endothermic, it absorbs heat.

H₂O_(l) ⇋ H⁺_(aq) + OH^-_(aq)     ΔH is positive
or, writing the chemical equation with ΔH as a reactant:
H₂O_(l) + ΔH ⇋ H⁺_(aq) + OH^-_(aq)

By Le Chatelier's Principle if the temperature of the system is increased, the equilibrium position will shift to the right to minimise the effect of the increased temperature.

The forward reaction is favoured resulting in the production of more H⁺_(aq) and more OH^-_(aq), so the value of K_w will increase (see K and Temperature).

Since the concentration of hydrogen ions is greater at higher temperatures, the pH of water will be lower at higher temperatures.

BUT, because the concentration of hydrogen ions in water is ALWAYS the same as the concentration of hydroxide ions in water, liquid water will ALWAYS be neutral!

The table below gives the value for the ion product for water, K_w, at various temperatures.
You should use these values to perform calculations to check the values given for [H⁺_(aq)] and pH.

Note that :

• The value of K_w at each of the temperatures above is very small so water undergoes very little self-dissociation.
  The percentage of dissociation (percent ionization) is small.
• As temperature increases, the value of K_w increases.
  That is, as temperature increases, production of products are favoured, so the reaction is endothermic.
• As temperature increases, the concentration of hydrogen ions increases.
  That is, as temperature increases, production of products are favoured, so the reaction is endothermic.
• As temperature increases, the pH of water decreases.
  BUT, the water does NOT become more acidic ....
• As the temperature increases, the water remains neutral!
  [H⁺_(aq)] = [OH^-_(aq)]