Question

Solubility of `Mg(OH)_2` is `1.6` x `10^-4` `"mol/L"` at `298` `K`. What is its solubility product?

Answer 1

`1.6 * 10^(-11)`

Explanation:

Magnesium hydroxide, `"Mg"("OH")_2`, is considered Insoluble in aqueous solution, which implies that the ions it forms will be in equilibrium with the solid.

Magnesium hydroxide dissociates only partially to form magnesium cations, `"Mg"^(2+)`, and hydroxide anions, `"OH"^(-)`

`"Mg"("OH")_text(2(s]) rightleftharpoons "Mg"_text((aq])^(2+) + color(red)(2)"OH"_text((aq])^(-)`

For an insoluble compound, its molar solubility tells you how many moles of the compound can be dissolved per liter of aqueous solution before reaching saturation.

In your case, a molar solubility of

`s = 1.6 * 10^(-4)`

means that you can only dissolve `1.6 * 10^(-4)` moles of magnesium in a liter of water at that temperature.

Take a look at the dissociation equilibrium. Notice that every mole of magnesium hydroxide that dissociates produces `1` mole of magnesium cations and `color(red)(2)` moles of hydroxide anions.

This tells you that if you successfully dissolve `1.6 * 10^(-4)` moles of magnesium hydroxide in one liter of solution, number of moles of each ion will be

`n_(Mg^(2+)) = 1 xx 1.6 * 10^(-4) = 1.6 * 10^(-4)"moles Mg"^(2+)`

and

`n_(OH^(-)) = color(red)(2) xx 1.6 * 10^(-4) = 3.2 * 10^(-4)"moles"`

Since we're working with one liter of solution, you can cay that

`["Mg"^(2+)] = 1.6 * 10^(-4)"M"`

`["OH"^(-)] = 3.2 * 10^(-4)"M"`

By definition, the solubility product constant, `K_(sp)`, will be equal to

`K_(sp) = ["Mg"^(2+)] * ["OH"^(-)]^color(red)(2)`

Plug in these values to get

`K_(sp) = 1.6 * 10^(-4) * (3.2 * 10^(-4))^color(red)(2)`

`K_(sp) = color(green)(1.6 * 10^(-11)) ->` rounded to two sig figs

The listed value for magnesium hydroxide's solubility product is `1.6 * 10^(-11)`, so this is an excellent result.

http://www.wiredchemist.com/chemistry/data/solubility-product-constants