Question

What mass of `"magnesium hydroxide"` is required to give a `1*L` volume of `[Mg(OH)_2]` whose concentration is `0.01*mol*L^-1` with respect to the hydroxide?

Answer 1

For a `1*L` volume of `0.01*mol*L^-1` `Mg(OH)_2`, `0.28*g` salt are required.

Explanation:

`"Molarity"` `=` `"Moles"/"Volume"` `=` `(0.01*mol)/(1.0*L)`.

And thus we need to dissolve `0.01*molxx58.32*g*mol^-1xx1/2` `=` `0.28*g` in a `1*L` volume. Why did I include the `1/2`?

However, `K_"sp",Mg(OH)_2=5.61×10^(−12)` at `298K`, which gives a solubility of `6.4xx10^-3*g*L^-1`. The question was thus not well-proposed.

Magnesium hydroxide is thus too insoluble to provide such a concentration.