[novice]

Hi everyone, just have a question that I have spent ages trying to solve without success.

What is the wight of Mg(2+) that must be added to 5 litres of a saturated Mg(OH)2 solution in order to lower the solubility of the Mg(OH)2 to 1/3 of its original value.

K(sp) of Mg(OH)2 at 25 degrees celcius = 1.40x10^-11
The original solubility of Mg(OH)2 i worked out to be 1.52x10^-4 M

I've tried manipulating K(sp)=[Mg2+][OH-]^2

Thankyou in advance!

PS: The answer is supposedly 0.160g

[Borek]

What will be concentration of OH^- when the solubility is three times lower than original?

[novice]

Is that an actual question, or is that the way to solve the problem, ascertain [OH-]

[Borek]

It's a hint.