# Using thermodynamics to predict the acid-base character of fluoride ion in water

I was working through the end-of-chapter exercises of the acid-base chapter in Shriver's Inorganic Chemistry when I came across the following problem:

4.7. The effective proton affinity $$\ce{A^{'}_{p}}$$ of $$\ce{F-}$$ in water is 1150 kJ $$\ce{mol^-}$$. Predict whether it will behave as an acid or base in water.

I was initially stumped by this problem since the $$\ce{A^{'}_{p}}$$ of $$\ce{F-}$$ is between the $$\ce{A^{'}_{p}}$$ values of water and hydroxide ion (1130 kJ mol- and 1188 kJ mol-), which tempted me to say $$\ce{F-}$$ is amphoteric in neutral water even though I know it's basic from general chemistry.

The answer that the book's solution manual provides is the following:

The proton affinities of H2O (l) and OH-(aq) are 1130 kJ mol- and 1188 kJ mol-, respectively. Consequently, the reaction below between $$\ce{H3O+}$$ and $$\ce{F-}$$ is exothermic by -20 kJ mol- (1130 kJ mol- minus 1150 kJ mol-):

$$\ce{ H3O+ (aq) + F- (aq) -> H2O (l) + HF (aq)}$$

The reaction below between $$\ce{H2O}$$ and $$\ce{F-}$$ is endothermic by 38 kJ mol- (1188 kJ mol- minus 1150 kJ mol-):

$$\ce{ H2O (l) + F- (aq) -> OH- (aq) + HF (aq)}$$

Therefore, in neutral water, $$\ce{F-}$$ will behave as an base.

I understand that the first reaction has an approximate enthalpy change of -20 kJ mol- and is favored because it is the reverse reaction of the following reaction between $$\ce{HF}$$ and $$\ce{H2O}$$ and reaction between acids and $$\ce{H2O}$$ are favorable when the $$\ce{A^{'}_{p}}$$ of $$\ce{H2O}$$ is greater than the $$\ce{A^{'}_{p}}$$ of the conjugate base:

$$\ce{ H2O (l) + HF (aq) -> H3O+ (aq) + F- (aq)}$$

However, I have a few points I don't understand:

1. Why does the text consider the second reaction with $$\ce{H2O}$$? Doesn't $$\ce{F-}$$ act as a base in both reactions?
2. If both reactions occur and $$\ce{F-}$$ acts as a base, then which one is dominant?
3. If the first reaction is dominant, then how does it proceed given the low concentrations ($$\ce{1 x 10^{-7} M}$$) of $$\ce{H3O+}$$ in neutral water?

The textbook walks us through the interpretation of these enthalpies (they also caution us not to use enthalpies as sole criterion for spontaneity of reactions):

So they are asking whether water reacts with the acid HA, and whether water reacts with the conjugate base $$\ce{A-}$$.

If we try that for the case at hand, we first consider whether HF reacts with water:

$$\ce{H2O(l) + HF(aq)->[?] H3O+(aq) + F−(aq)}$$

Here, water and fluoride are the bases "fighting for the proton" (reactants: fluoride has the proton; products: water has the proton). Fluoride wins (i.e. HF does not give away the proton) because it has the higher proton affinity, and the reaction does not proceed judged by enthalpy alone.

Second, we consider whether the fluoride ion reacts with water:

$$\ce{H2O(l) + F-(aq) ->[?] OH-(aq) + HF(aq)}$$

Here, fluoride and hydroxide are the bases "fighting for the proton". Hydroxide wins because it has the higher proton affinity, i.e. this reaction also does not proceed judged by enthalpy alone.

So HF is not a strong acid, and fluoride is not a strong base. Another way of saying this is that hydroxide is a stronger base than fluoride, and hydronium is a stronger acid than HF.

In the textbook walk-through, they conclude that HA is a strong acid and - for a different reaction - $$\ce{A-}$$ is a strong base. In our case, HF is a weak acid and fluoride is its conjugate weak base. If you try the same question with HCl, you will see it is a strong acid and chloride is not a strong base. (I don't understand the conclusion of the solution manual - they might have gotten confused because they know fluoride is a weak base, but their calculation together with the textbook argument merely shows that it is not a strong base.)

1. Why does the text consider the second reaction with H2O? Doesn't F− act as a base in both reactions?

I think they ask "does HF react with water?", and "does fluoride react with water?". I don't know why the answer flips one of the reactions (maybe the solution manual wasn't written by the textbook author, and the original intent of the question got lost).

1. If both reactions occur and F− acts as a base, then which one is dominant?

There are three equilibrium reactions, the two shown, and auto-dissociation of water. I don't know the kinetics of the three, but they are all fast and the three equilibrium constants are linked because if you subtract the two reactions you show, you get the auto-dissociation of water. For the equilibrium state, it does not matter which set of two you consider.

1. If the first reaction is dominant, then how does it proceed given the low concentrations (1x 10−7 M) of H3O+ in neutral water?

Reaction rate constants vary by many orders of magnitude. The reaction of fluoride and hydronium is nice because the reactants are of opposite charge, so maybe it has a high rate constant. The reaction of fluoride and water is nice because water is the solvent, so it is present at high concentration and already at close proximity. I could not tell you which reaction is faster even if I know the concentrations of all the reactants.

Here is the table of proton affinities as a quick reference: