# Why are the reactants in the solubility product constant multiplied by their coefficients and raised to the power of their coefficients?

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?

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 $(2X)^2$.
• 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? – Curt F. Aug 10 '15 at 3:21