# Calculation of equilibrium constant for dissolution of iron(II) fluoride in acid

I am reviewing for the AP Chemistry test, and the question is this

$$\ce{FeF2(s) <=> Fe^2+(aq) + 2F^- (aq) } \quad K_1 = 2\times 10^{-6}$$

$$\ce{F- (aq) + H+ (aq) <=> HF(aq)} \quad K_2 = 1 \times 10^3$$ $$\ce{FeF2 (s) + 2 H+ <=> Fe^2+ (aq) + 2 HF (aq)} \quad K_3 = ?$$

I solved for $$K_3$$ by multiplying $$K_1$$ and $$K_2$$, and then dividing by 2. I got $$K_3 = 1 \times 10^{-3}$$. I would think that this would be not thermodynamically favorable since $$K_3 < 1$$, but the answer was the opposite: $$K_3$$ is thermodynamically favorable, because $$K_3 > 1$$. Why is this the answer?

• I think (but I should have really gone to bed three hours ago) that you should have multiplied the value but squared $K_2$ (you need that equation twice) because thermodynamics. I’m not $100~\%$ sure though. Oh yeah: Please don’t use MathJax in titles ;) – Jan May 1 '16 at 1:58
• Ah, that might be the answer. My reasoning for the division by two was using $K = \frac{[HF]}{[H][F]}$, and replacing each one with twice the amount. – Blubber May 1 '16 at 2:05

Let’s approach the entire problem from the other direction. Consider your final reaction:

$$\ce{FeF2(s) + 2H+ (aq) <=> Fe^2+ (aq) + 2 HF (aq)}\\ K_\mathrm{tot} = \frac{[\ce{Fe^2+}][\ce{HF}]^2}{[\ce{FeF2}][\ce{H+}]^2}$$

We have our initial dissociation with the following equation:

$$\ce{FeF2(s) <=> Fe^2+ (aq) + 2 F- (aq)}\\ K_1 = \frac{[\ce{Fe^2+}][\ce{F-}]^2}{[\ce{FeF2}]}$$

And of course we have the second equilibrium which equates to:

$$\ce{F- (aq) + H+ (aq) <=> HF (aq)}\\ K_2 = \frac{[\ce{HF}]}{[\ce{F-}][\ce{H+}]}$$

Now we want to express $K_\mathrm{tot}$ in terms of $K_1$ and $K_2$. Since we are dealing with products all the way through, it will somehow be a multiplication. So let’s try $K_1 \times K_2$:

$$K_1 \times K_2 = \frac{[\ce{Fe^2+}][\ce{F-}]^2}{[\ce{FeF2}]} \times \frac{[\ce{HF}]}{[\ce{F-}][\ce{H+}]} = \frac{[\ce{Fe^2+}][\ce{F-}][\ce{HF}]}{[\ce{FeF2}][\ce{H+}]}$$

We’re not quite there yet. We still have an extraneous fluoride concentration in the numerator and we are missing an $\ce{HF}$ concentration and a proton concentration. Luckily, that is exactly what an additional $K_2$ will give us:

$$K_1 \times \left( K_2 \right )^2 = \frac{[\ce{Fe^2+}][\ce{F-}]^2}{[\ce{FeF2}]} \times \left ( \frac{[\ce{HF}]}{[\ce{F-}][\ce{H+}]} \right )^2 = \frac{[\ce{Fe^2+}][\ce{HF}]^2}{[\ce{FeF2}][\ce{H+}]^2} = K_\mathrm{tot}$$

This is the mathematical way of saying ‘we need the second equation twice’. Due to the way equilibrium constants work, you always have to multiply or divide, so ‘doing something twice’ is equivalent to squaring.

Numerically, this means the answer is:

$$K_\mathrm{tot} = K_1 \times \left ( K_2 \right )^2 = 2 \times 10^{-6} \times \left ( 1 \times 10^3 \right )^2 = 2 \times 10^{-6} \times 1 \times 10^6 = 2$$

Yes, $\ce{FeF2}$ is a solid, so it has an activity of $1$ so we could ignore it when calculating the equilibrium constant. But we don’t actually use its value, so its being there doesn’t hurt either.

Yes, Jan is right. You need to multiply $K_2$ squared because it takes two additions of reaction 2 to get from reaction 1 to reaction 3. You will get $K_3$ = 2.

As a side note, and you probably meant this anyway, but saying that a molecule is "thermodynamically favorable" is meaningless. You should say "the dissolution of [molecule] in [condition] is thermodynamically favorable." And keep in mind that thermodynamically favorable in standard conditions refers to just that: standard conditions--1 molar solutions. Thus, even if K was less than 1, you could still say that it is thermodynamically favorable for [a certain small amount] to dissolve."

• Yes, that is what I meant and it was the original wording of the question was. I omitted the details of the question because the bottom left corner says "Unauthorized copy or reuse of any part of this page is illegal"....but hopefully my educational purpose won't put me in jail. – Blubber May 1 '16 at 2:12