Question

What is the `"pH"` of a solution of magnesium hydroxide given that `K_(sp) = 1.8 * 10^(-11)` ?

Answer 1

`"pH" = 10.52`

Explanation:

In order to find the pH of a solution, you must determine the concentration of hydronium ions, `"H"_3"O"^(+)`, either directly or indirectly.

When you're dealing with a Bronsted - Lowry acid, you will be solving for the concentration of hydronium ions directly.

When you're dealing with a Bronsted - Lowry base, like you are here, you will be solving for the concentration of hydronium ions indirectly, i.e. by solving for the concentration of hydroxide ions, `"OH"^(-)`.

Magnesium hydroxide, `"Mg"("OH")_2`, us insoluble in aqueous solution. This means that when you place magnesium hydroxide in water, an equilibrium will be established between the undissolved solid and the dissolved ions.

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

The solubility product constant, `K_(sp)`, for this equilibrium looks like this

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

What you need to do here is use the `K_(sp)` to find the molar solubility, `s`, of magnesium hydroxide in aqueous solution at `25^@"C"`.

To do that, set up an ICE table

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

`color(purple)("I")" " " " " "-" " " " " " " " " " " " " "0" " " " " " " " " " 0`
`color(purple)("C")" " " " " "-" " " " " " " " " " " "(+s)" " " " " "(+color(red)(2)s)`
`color(purple)("E")" " " " " "-" " " " " " " " " " " " " "s" " " " " " " " " "color(red)(2)s`

Remember, the concentration of the solid is assumed to be constant.

Here `K_(sp)` will be equal to

`K_(sp) = s * (color(red)(2)s)^color(red)(2) = 4s^3`

Rearrange to solve for `s`

`s = root(3)(K_(sp)/4) = root(3)((1.8 * 10^(-11))/4) = 1.65 * 10^(-4)`

This means that the concentration of hydroxide ions in a saturated solution of magnesium hydroxide at `25^@"C"` will be

`["OH"^(-)] = color(red)(2) * 1.65 * 10^(-4)"M" = 3.30 * 10^(-4)"M"`

Now, at `25^@"C"`, the concentration of hydronium ions and the concentration of hydroxide ions have the following relationship

`color(blue)(K_W = ["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-4))" "`, where

`K_W` - the ion product for water's auto-ionization

Plug in your value to get

`["H"_3"O"^(+)] = 10^(-14)/(3.30 * 10^(-4)) = 3.03 * 10^(-11)"M"`

This means that the pH of the solution will be

`color(blue)("pH" = - log(["H"_3"O"^(+)]))`

`"pH" = - log(3.03 * 10^(-11)) = color(green)(10.52)`

Alternatively, you can use the equation

`color(blue)("pH " + " pOH" = 14)`

Here

`color(blue)("pOH" = - log(["OH"^(-)]))`