On the activity of liquids and solvents in equilibium constants

This is a confusion I've had for quite sometimes, first when learning about equilibria (due to contradicting teachers), then when trying to teach it (due to contradicting sources, in books and on the Internet).

When facing an equilibrium (let's be original and take $\ce{a A + b B <=> c C + d D}$ as an example), the equilibrium constant is $$K = \frac{a(\ce{C})^c \cdot a(\ce{D})^d}{a(\ce{A})^a \cdot a(\ce{B})^b}$$ where $a(\ce{X})$ is the activity of X.

For pure solids, the activity is 1; therefore solids "do not appear" in the expression of the constant. For dilute solutions, the activity of the solute can be taken as its molar concentration. Thus, supposing A, B, C and D are not solids, the equilibrium constant can be written as

$$K_c = \frac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$

But what happens when facing liquids (e. g. water)? In that case, I have seen two things :

1. Just putting the "concentration" of the liquid into the equilibrium constant.

2. A pure (or nearly pure, such as the solvant in a dilute solution) liquid has an activity of (or nearly of) 1. Therefore, the liquid should not appear in the constant.

Let's apply both cases to acidity constants, for the reaction $\ce{HA + H2O <=> A- + H3O+}$.

1. As $[\ce{H2O}] \approx 55.55$ mol/l in dilute solutions, $$K_c = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}][\ce{H2O}]} \textrm{ and } K_a = \frac{K_c}{[\ce{H2O}]} = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$
2. As $a(\ce{H2O}) \approx 1$ in dilute solutions, $$K_a = K_c = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$

Which way is correct? I'd be more inclined to follow the option #2, but I've been stared at wide-eyed like a heretic when talking about that.

I concluded on this question after discussion with the physical chemistry professor at my home university, and reading some physical chemistry books.

Although many references include the concentration of water in the expression of Ka, this is incorrect: the activity of pure water is 1. In dilute solutions, $$a(\ce{H2O}) \approx 1$$. Therefore, the second option is correct: $$K_a=K_c=\frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$

This has a direct consequence on the values of pKa for the two couples of water: $$pKa(\ce{H3O+}/\ce{H2O})=0$$ and $$pKa(\ce{H2O}/\ce{OH-})=14$$, instead of the values $$-1.7$$ and $$15.7$$ that are often found in textbooks (which result from including $$[\ce{H2O}]=55.55$$ mol/l).

A good article on this matter can be found here: What is the pKa of water?.

References:

• B. Leyh, Department of Chemistry, University of Liège, Belgium.
• Atkins, P., & De Paula, J. (2006). Atkins’ physical chemistry (8th ed.). London, England: Oxford University Press.
• Reid, P., & Engel, T. (2006). Physical Chemistry. Philadelphia, PA: Pearson Education.

You should go through the derivation of the equilibrium constant from thermodynamics and see exactly what is meant by activity, and the importance of defining a "standard state". The activity of water in a dilute solution is very nearly 1 because our standard state assumes pure water. We use the concentration of solutes because the standard state typically involves 1M solutions. Using the concentration of water is wrong unless you define your standard state with 1M water (in...something?).

If you want to be extra careful, you might write the activity of, for example, C in your first equilibrium as $$\left( \frac{[C]}{[C]°} \right)^c$$ Where the little ° means "in the reference state".

• I am well aware of the definition of the activity, and that's where my problem is: there seems to be no consensus on what the value of $[\ce{H2O}]°$ is. Shouldn't there be a standardized standard state for water? Commented Jul 25, 2017 at 14:21