Calculation of equilibrium constant for dissolution of iron(II) fluoride in acid
Let’s approach the entire problem from the other direction. Consider your final reaction:

$$\ce{FeF2(s) + 2H+ (aq) &lt;=&gt; Fe^2+ (aq) + 2 HF (aq)}\\
K_\mathrm{tot} = \frac{[\ce{Fe^2+}][\ce{HF}]^2}{[\ce{FeF2}][\ce{H+}]^2}$$

We have our initial dissociation with the following equation:

$$\ce{FeF2(s) &lt;=&gt; Fe^2+ (aq) + 2 F- (aq)}\\
K_1 = \frac{[\ce{Fe^2+}][\ce{F-}]^2}{[\ce{FeF2}]}$$

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

$$\ce{F- (aq) + H+ (aq) &lt;=&gt; HF (aq)}\\
K_2 = \frac{[\ce{HF}]}{[\ce{F-}][\ce{H+}]}$$

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.