1.
Mg (OH)2 <-----> Mg2+ + 2OH-
                                        s               2s
Ksp = [Mg2+] [OH-]^2
7.1*10^-12 = s * (2s)^2
7.1*10^-12 = 4*s^3
s = 1.21*10^-4 M
[OH-] = 2s
              = 2* 1.21*10^-4
              = 2.42*10^-4 M
Answer: 2.42*10^-4 M

2.
Mg(OH)2 dissolved = 1.21*10^-4 mL
V = 273.5 mL = 0.2735 L
number of moles of Mg(OH)2 = M*V
                           = 1.21*10^-4 * 0.2735
                           = 3.31*10^-5 mol
So,
amount of Mg(OH)2 dissolved = number of moles * molar mass
                                                               = (3.31*10^-5) * 58.32
                                                               = 1.931*10^-3 g
                                                                = 1.931 mg
Amount of Mg(OH)2 to be recovered = 361.824 mg - 1.931 mg = 359.893 mg
Answer: 359.893 mg

only 1 question at a time please