If a solution of magnesium hydroxide Mg(OH)2 has a concentration of 1.8E-6 M, calculate the concentration of its hydroxide ion.

Magnesium hydroxide is a weak base with low solubility in water. It contains two hydroxide ions (OH-) per formula unit, so the number of hydroxide ions per liter in 1.8 x 10E-6 M solution is:

2(1.8 x 10E-6) = 3.6 x 10E-6.

However, this doesn't technically answer the question, as most of these hydroxide ions will not have dissociated from the magnesium ions due to the compound's low solubility. To calculate the concentration of dissociated OH- ions in solution we need to know the Ksp (solubility product constant) for magnesium hydroxide. This value is 5.6 x 10E-12 at 25ºC.

Here's what the Ksp value means: The amount of magnesium hydroxide that dissolves will be such that the product of the magnesium ion and hydroxide ion concentrations in solution, raised to the power of their coefficients, will be equal to 5.6 x 10E-12:

`Mg(OH)_2 -> Mg^(2+) + 2 OH^-`

`[Mg^(2+)][OH^-]^2 = K_(sp) = 5.6 x 10^(-12)`

Since there are twice as many dissolved hydroxide ions as magnesium ions (based of the formula of magnesium hydroxide) we can say that [Mg(2+)] = X and [OH-] = 2X, then solve for [OH-]:

`(X)(2X)^2 = 5.6 x 10^(-12)`

`4X^3 = 5.6 X 10^(-12)`

`X = 1.12 x 10^(-4)`

`[OH-] = 2X = 2.22 x 10^(-4) M`