Question

1. We define the activity of pure liquids as 1 to define the standard state. However, this is an arbitrary choice. Find the dissociation constant, K, for the autoprotolysis of pure water using the molarity standard state for water (i.e. defining the activity as 1 for a water concentration of 1 mol/l). You may take the density of water to be 1.0 g/ml.

2.   Will K have the same value in this standard state at 60?C?   

Answer 1

• STANDERED STATE MEANS
• TEMPERATURE = 298.15 K
• VOLUME = 22.4 LITRE
• PRESSURE= 1 ATM
• AUTOPROTOLYSIS OF PURE WATER GENERATE EQUAL AMOUNT OF [H]+ AND[OH]^-
• H₂O  [H]⁺ + [OH]^-
• DISSOCATION CONSTANT OF PURE WATER IS IONIC PRODUCT OF PURE WATER
• ^K_W =[H]⁺[OH]^- = 10^-14
• [H]⁺ = [OH]^-   = 10^-7
• 333-15 k temperature means temperature increase.
• The formation of [H]⁺ and [OH]^- from water is an endothermic process.

• According to Le Châtelier's Principle if reaction conditions are change in dynamic equilibrium, the position of equilibrium moves to counter the change you have made.

• Hence, if you increase the temperature of the water, the equilibrium will move to lower the temperature again. It will do that by absorbing the extra heat.

• That means that the forward reaction will be favored, and more hydrogen ions and hydroxide ions will be formed.

• The effect of that is to increase the value of Kw] as temperature increases.