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It is probably easiest for me to explain the question in an example. Consider the following chemical reaction:

$\ce{MgF2(s) -> Mg^{2+} (aq) + 2F- (aq)}$

Plugging this into the formula for the solubility product constant, $K_{sp}=[M^{y+}]^{x}[A^{x-}]^{y}$, we get:

$K_{sp}=[\ce{Mg^{2+}}][\ce{F-}]^{2}$.

The solubility product constant for magnesium fluoride is $5.16\times10^{-11}$. To determine molar solubility, the equation to solve for magnesium and fluoride is $5.16\times10^{-11}=X\times (2X)^{2}$.

$X$ is magnesium and $(2X)^{2}$ is fluoride. This is because there are two moles of fluoride for every one mole of $\ce{MgFl_{2}}$, and one mole of magnesium for every one mole of $\ce{MgFl_{2}}$. (Please correct me if I'm wrong anywhere.)

But here's whats confusing to me. Why is fluoride denoted as $(2X)^{2}$? I know that there are two moles of fluoride per one mole of magnesium fluoride, but why does it need to be denoted by being both multiplied to its coefficient and raised to the power of that coefficient?

Sources: Magnesium Fluoride Solubility product constant

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The reason why in the equilibrium constant expression, the concentration is to the power of coefficient is because the above equation can be written like this: $$\ce{MgF_2(s) -> Mg^{2+} (aq) + F^- + F^- (aq)}$$ Hence when you plug this into the equilibrium constant expression, you get this: $$K_{sp}=[\ce{Mg^{2+}}][\ce{F-}][\ce{F-}]$$ which becomes the equivalent to:$$K_{sp}=[\ce{Mg^{2+}}][\ce{F-}]^{2}$$ So really the equilibrium constant is proportional to the square of the concentration of fluoride. So that is the reason for the square. However, the concentration of fluoride is twice as much as the concentration of Mg, hence that is the reason for the multiplying the concentration of fluoride by 2. So that is why fluoride is denoted as $(2X)^2$.

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  • $\begingroup$ How is the fluoride concentration twice as much? I understand why the equilibrium constant expression squares, but why would it multiply? It seems redundant. $\endgroup$ – Isaac Liu Aug 9 '15 at 18:16
  • $\begingroup$ The reason the fluoride concentration is squared is due to the equilibrium constant. The reason why the fluoride concentration is also multiplied by two is because of its relationship with Mg $\endgroup$ – Nanoputian Aug 9 '15 at 20:27
  • $\begingroup$ Say you have $X$ moles of magnesium fluoride powder that you dissolve in water. After it dissolves, how many moles of fluoride anion will there be in solution? $\endgroup$ – Curt F. Aug 10 '15 at 3:21
  • $\begingroup$ We'd get 2X moles of fluoride anion. Thanks for both answers, that clears it up. Raising it to the square has nothing to do with its relationship with Mg, it is just part of the equation. $\endgroup$ – Isaac Liu Aug 10 '15 at 21:33

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