# What's the point of solubility products?

What is the advantage of solubility products, e.g.

LiF(s) ⇄ Li+(aq) + F-(aq) | K = [Li+(aq)] * [F-(aq)] = 0.00184 M2 at 25°C (source)

compared to just stating that

Solubility in water = 0.134 g/100 mL (25 °C) (source)

(There's a little difference between (0.00184 M2)½ = 1.11 g/L and the solubility of 1.34 g/L above; I take it that's because of the difference between activity and concentration?)

It seems to me that solubility products are just a roundabout, complicated way of expressing the concentration of a saturated solution. I suspect that I'm missing something though: otherwise why would people publish long lists of solubility products?

What I read: Wikipedia; all my high school textbooks.

• It all starts making sense when you are dealing with a mixture of salts having a common ion. – Ivan Neretin May 27 '18 at 10:26
• Oh OK. In a situation like, how much LiF can I dissolve in a solution that already contains 30 g/L NaF; or what is the concentration of Na+, Li+ and F- in a saturated solution of LiF and NaF. (I'd have to think about how to solve the resulting equations...) – Martin May 27 '18 at 10:57
• Yeah, that's exactly what is it good for. – Ivan Neretin May 27 '18 at 11:11

Solubility products figure prominently in equilibria involving precipitated species. Following is an example that may arise in a practical situation.

Problem: We have 0.01 M ferrous ion in water and we propose adding a base to drive out the iron as hydroxide. We do not want excess base dissolving into the water so we will try magnesium hydroxide. How well will it work?

The proposed reaction is then

$\ce{Fe^{2+} + Mg(OH)2(s) <=> Mg^{2+} + Fe(OH)2(s)}$

$K=\frac{[\ce{Mg^{2+}}]}{[\ce{Fe^{2+}}]}$

Compare this equilibrium constant with:

$K_{sp}(\ce{Mg(OH)2})=[\ce{Mg^{2+}}][\ce{OH^-}]^2=5.61×10^{-12}$ (source)

$K_{sp}(\ce{Fe(OH)2})=[\ce{Fe^{2+}}][\ce{OH^-}]^2=4.87×10^{-17}$ (Ibid.)

So then

$K=\frac{[\ce{Mg^{2+}}]}{[\ce{Fe^{2+}}]}=\frac{K_{sp}(\ce{Mg(OH)2})}{K_{sp}(\ce{Fe(OH)2})}=1.15×10^5$

Since the reaction stoichionmetry implies that the sum $[\ce{Mg^{2+}}]+[\ce{Fe^{2+}}]$ will be constant at the original iron concentration of 0.01 M, we find that the iron level remaining in the water may be reduced by five orders of magnitude, the equilibrium concentration is below $10^{-7}$ molar!

Magnesium hydroxide, despite being supposedly "weak" because of its limited solubility, is actually like a strong base here, displacing the weak iron hydroxide base almost quantitatively -- but, in effect, self-regulating because of that limited intrinsic solubility. The same property applies with respect to most heavy metals commonly found in water; their hydroxide solubilities, like that of iron, are much less than magnesium's. Thus magnesium hydroxide (or, in practice, magnesium oxide which becomes the hydroxide in situ) is a good agent for treating water to remove heavy metals. See here, under "Applications".