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### The Curious Case of Water in Equilibrium Constant Expressions

posted Dec 7, 2016, 12:24 AM by Grace Ong   [ updated Dec 7, 2016, 12:24 AM ]
The following post was first posted on Blogger on Friday, 19 July 2013.

A student and I recently had an interesting discussion on a rather perplexing problem − should we or should we not include the concentration water in equilibrium constant expressions?

This is a my favourite scene from Disney's beloved classic, Alice in Wonderland.
And no, this is not the Johnny Depp version.

Why is water appearing and disappearing from equilibrium constant expressions like the Cheshire Cat,
teasing and confusing my dear students?

We were discussing Question 2(b)(iv) from the Nov 2010 Paper 3, in which students were asked to write an expression for Kc, and calculate its value for the following equilibrium by assuming that [H2O] = 55.5 mol dm−3 throughout.

As my student was taught to ignore water concentrations for reactions taking place in aqueous media, she calculated the equilibrium constant using the following expression:

Unfortunately for her (and for many other Singapore candidates in 2010), this was not what the Cambridge examiners were expecting. Since water was part of the reaction, examiners were expecting candidates to include the water concentration as part of the Kc expression, i.e.

That was why the value of the water concentration was given in the question − to be used in the calculations!

As discussed in my previous post, the concentration of water is generally omitted from the equilibrium constant expression for an aqueous equilibrium, even if water takes part in the equilibrium as a reactant or a product. However, I also cautioned that this omission should not be applied across the board without first giving due consideration to the purpose for which the equilibrium constant expression is defined.

In the case of the GDL-gluconic acid equilibrium, the question asked for the Kc expression, not an expression in which the water concentration was subsumed under a re-defined equilibrium constant. Students were thus required to include the concentrations of all reactants and products in strict accordance with the Equilibrium Law. That the value of the water concentration was given in the question should be a clue to students that this was the requirement.

Let's consider two more equilibria taken from past examination papers.

 E.g. 1 Nov 2002 Paper 1 Question 8 H2O(l) ⇌ H+(aq) + OH−(aq)

How should we write the Ka expression for this equilibrium?

Now, if we simply omit the concentration of water, the resulting 'Ka' expression would look exactly like the ionic product of water, Kw = [H+][OH], which can't be right!

Ka, the acid dissociation constant, measures quantitatively the extent to which acid molecules dissociate to form (H+) ions in solution. The stronger the acid, the greater the amount of H+ ions formed in relation to the amount of undissociated acid molecules. By this definition, we cannot omit the concentration of undissociated acid molecules from the Ka expression.

For a weak acid, HA, that dissociates in water as follows:

HA(aq) + H2O(l) ⇌ A(aq) + H3O+(aq)
or in simplified form, HA(aq) ⇌ A(aq) + H+(aq)

To write a Ka expression for H2O, we regard H2O as a weak acid HA, i.e. we can write it as HOH,

HOH(l) + H2O(l) ⇌ OH(aq) + H3O+(aq)
or in simplified form, HOH(l) ⇌ OH(aq) + H+(aq)

Thus the Ka of water should be written as follows:

Indeed, the concentration of water was given in the question as mol dm−3, and students were expected to use this value (and the given equilibrium concentration of H+ ions at 25 °C) to calculate the Ka of water.

Note:
1. Those who are sharp will notice that there IS one water concentration omitted here - that of water functioning as a base. In the definition of Ka, the concentration of water (functioning as a base) is omitted since it is in large excess as compared to the concentration of the weak acid, HA.
2. For a detailed explanation, please refer to my earlier post which explains the relationship between Kc and Ka.
3. The base dissociation constantKb, of water can be similarly derived by regarding H2O as a weak base B:

B(aq) + H2O(l) ⇌ BH+(aq) + OH(aq)
H2O(l) + H2O(l) ⇌ H2O−H+(aq) + OH(aq)

4. The Ka of water is exactly equal in value to the Kb of water. At 25 °C, each has a value of 1.8 × 10−16 mol dm−3, since [H2O] = mol dm−3, and [H+(aq)] = [OH(aq)] = 10−7 mol dm−3 at this temperature.

 E.g. 2 Nov 2005 Paper 2 Question 1(c) D2O(l) + H2O(l) ⇌ 2HDO(l)     where D = deuterium

What is the equilibrium constant expression for this reaction?

There are two ways to approach this problem.
• First, this is a homogeneous equilibrium, which means that the concentrations of all reactants and products must be included in the equilibrium constant expression, as dictated by the Equilibrium Law.
• Second, comparable initial masses of D2O and H2O were used in this reaction, which meant that none of the reactant concentrations could be assumed to be approximately unchanged after equilibrium is reached. Thus we cannot re-define another equilibrium constant expression by subsuming any of the concentrations under a new constant.
Therefore, the Kc expression for this reaction is