The Ksp of Mg(OH)2 in water is 1.8 x 10-<span>11. This means that in pure water, Mg(OH)2 has a solubility of:
</span>∛[(1.8 x 10-11) / 4] = 1.65 x 10-4 mol/L
<span>which is equal to
</span>1.65 x 10-4<span> mol x (58.32) / 10 x 100 mL = 9.62 x 10-4g / 1x102 mL
If the pH is 12, the hydroxide concentration in the solvent is
10^-(14-12) = 0.01 mol/L
The solubility is solve using the formula
</span>1.8 x 10-11 = x (2(0.01 + x))^2
x = 4.5x10-8 mol/L
which is equal to
4.5x10-8 mol x (58.32) / 10 x 100 mL = 2.62 x 10-7g / 1x102 mL
Answer:

Explanation:
Hello!
In this case, since the given reaction is:

Whereas the equilibrium constant is:
![K=\frac{[C][D]^{5/2}}{[A][B]^2} =4.0](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BC%5D%5BD%5D%5E%7B5%2F2%7D%7D%7B%5BA%5D%5BB%5D%5E2%7D%20%3D4.0)
However, the new target reaction reverses and doubles the initial reaction to obtain:

Whereas the equilibrium constant is:
![K'=\frac{[A]^2[B]^4}{[C]^2[D]^5}](https://tex.z-dn.net/?f=K%27%3D%5Cfrac%7B%5BA%5D%5E2%5BB%5D%5E4%7D%7B%5BC%5D%5E2%5BD%5D%5E5%7D)
Which suggest the following relationship between the equilibrium constants:

So we plug in to obtain:

Best regards!
There are two elements, two hydrogen to be exact.