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

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

Summary Submit Skip Question

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.