Answer:
![pH=12.3\\\\pOH=1.7\\](https://tex.z-dn.net/?f=pH%3D12.3%5C%5C%5C%5CpOH%3D1.7%5C%5C)
![[H^+]=5x10^{-13}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D5x10%5E%7B-13%7DM)
![[OH^-]=0.02M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.02M)
Explanation:
Hello there!
In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:
![Mg(OH)_2(s)\rightleftharpoons Mg^{2+}(aq)+2OH^-(aq)](https://tex.z-dn.net/?f=Mg%28OH%29_2%28s%29%5Crightleftharpoons%20Mg%5E%7B2%2B%7D%28aq%29%2B2OH%5E-%28aq%29)
Thus, since the ionization occurs at an extent of 1/3, we can set up the following relationship:
![\frac{1}{3} =\frac{x}{[Mg(OH)_2]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%20%3D%5Cfrac%7Bx%7D%7B%5BMg%28OH%29_2%5D%7D)
Thus, x for this problem is:
![x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x= 0.01M](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%5BMg%28OH%29_2%5D%7D%7B3%7D%3D%5Cfrac%7B0.03M%7D%7B3%7D%5C%5C%5C%5Cx%3D%20%200.01M)
Now, according to an ICE table, we have that:
![[OH^-]=2x=2*0.01M=0.02M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D2x%3D2%2A0.01M%3D0.02M)
Therefore, we can calculate the H^+, pH and pOH now:
![[H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5Cfrac%7B1x10%5E%7B-14%7D%7D%7B0.02%7D%3D5x10%5E%7B-13%7DM)
![pH=-log(5x10^{-13})=12.3\\\\pOH=14-pH=14-12.3=1.7](https://tex.z-dn.net/?f=pH%3D-log%285x10%5E%7B-13%7D%29%3D12.3%5C%5C%5C%5CpOH%3D14-pH%3D14-12.3%3D1.7)
Best regards!