Hey, Can someone please explain to me how to do part D.

Ksp= 1.41 x 10^-12

Thanks!

Hey there,

There are a few steps involved, but hopefully I'm making some sense!

1. Write an equation (as always)

Note the molar ratios - the OH has 2.

2. Write a K

_{sp} expression:

That's the solubility product, considering the OH's molar ratio. We don't divide by the magnesium hydroxide as that's a solid when it isn't dissolved.

3. Replace the concentration of Mg

^{2+} with x. We know this to be the same concentration as that of magnesium hydroxide (what we're trying to figure out).

4. Solving for x gives: 7.06 *10

^{-5}M. This is the molar solubility of Mg(OH)

_{2} - the amount of moles that can be dissolved in a litre of water.

An aside: regular solubility is measured in g/L. To convert, just multiply 7.06 *10

^{-5} by the molar mass of Mg(OH)

_{2}