Question

If a weak acid is say only 5% ionized at equilibrium, then the ionization rxn would be reactant favored, correct? I am a little on the fence about understanding this if anyone could better explain why this would be? I am assuming because it is not completely ionized, the amount of products would be less than the amount of reactants and so it would be reactant favored.

Answer 1

Here is the way I do it.
1. Write the reaction. Let's use acetic acid, CH3COOH (which I will call HAc), as an example.
HAc ==> H^+ + Ac^-

2. Write the Ka expression. For a weak base write the Kb expression. For an equilibrium reaction, write Keq. They all work the same.
Ka = 1.8 x 10^-5 = (H^+)(Ac^-)/(HAc).

3. Analyze.
It's as simple as this. Ka is small; therefore, the numerator of the fraction must be smaller than the denominator. Which means, of course, that the ions are less than the un-ionized material and that means the products are not favored (the reactants are favored). I look at it that the forward reaction is not favored; the reverse reaction is favored. (Another way I find useful is to say, "not many ions, a lot of un-ionized acid")

4. Students always want to know what I consider small; i.e., what Ka, Kb, Keq, is small and what is large. For a 1:1 set of coefficients, a Ka, Kb, Keq of 1 means numerator of fraction = denominator which means the reaction is 50%. So I use 1 as a benchmark although reactions that are not 1:1 vary slightly but as a quick assessment, K of 1 is a good number to choose.