$$\ce{FeF2(s) <--> Fe^{2+}(aq) + 2F-(aq)} \hskip{2em} K_1=2\times 10^{-6}$$ $$\ce{F-(aq) + H+(aq) <--> HF(aq)} \hskip{2em} K_2=1\times 10^3$$ $$\ce{FeF2(s) + 2H+(aq) <--> Fe^{2+}(aq) + 2HF(aq)} \hskip{2em} K_3=?$$

On the basis of the information above, the dissolution of $\ce{FeF2(s)}$ in acidic solution is

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The dissolution of $\ce{FeF2(s)}$ in acidic solution is represented by the last reaction. We can determine if it is thermodynamically favored by finding the value of $K_3$.

The following equilibrium equations can be written for the first two equations (remember that solids are not included in equilibrium equations).

$$\ce{FeF2(s) <--> Fe^{2+}(aq) + 2F-(aq)} \hskip{2em} K_1=2\times 10^{-6}$$ $$K_1 = [\ce{F-}]^2[\ce{Fe^{2+}}]$$ $$\ce{F-(aq) + H+(aq) <--> HF(aq)} \hskip{2em} K_2=1\times 10^3$$ $$K_2 = \frac{[\ce{HF}]}{[\ce{H+}][\ce{F-}]}$$

We can also write an equilibrium equation for the third reaction:

$$K_3 = \frac{[\ce{Fe^{2+}}][\ce{HF}]^2}{[\ce{H+}]^2}$$

$K_3$ can be obtained from $K_1$ and $K_2$ the following way:

$$K_3 = K_1\cdot K_2^2$$ $$\frac{[\ce{Fe^{2+}}][\ce{HF}]^2}{[\ce{H+}]^2} = [\ce{F-}]^2[\ce{Fe^{2+}}]\cdot\left(\frac{[\ce{HF}]}{[\ce{H+}][\ce{F-}]}\right)^2$$ $$\frac{[\ce{Fe^{2+}}][\ce{HF}]^2}{[\ce{H+}]^2} = [\ce{F-}]^2[\ce{Fe^{2+}}]\cdot \frac{[\ce{HF}]^2}{[\ce{H+}]^2[\ce{F-}]^2}$$ $$\frac{[\ce{Fe^{2+}}][\ce{HF}]^2}{[\ce{H+}]^2} = \cancel{[\ce{F-}]^2}[\ce{Fe^{2+}}]\cdot \frac{[\ce{HF}]^2}{[\ce{H+}]^2\cancel{[\ce{F-}]^2}}$$ $$\frac{[\ce{Fe^{2+}}][\ce{HF}]^2}{[\ce{H+}]^2} = \frac{[\ce{Fe^{2+}}][\ce{HF}]^2}{[\ce{H+}]^2}$$

Therefore,

$$K_3 = K_1\cdot K_2^2$$ $$K_3 = (2\times 10^{-6}) \cdot (1\times 10^3)^2$$ $$K_3 = (2\times 10^{-6})\cdot(1\times 10^6)$$ $$K_3 = 2$$

Since $K_3 \gt 1$, the forward reaction is favored.