What is the pH of a 0.1 M solution of each of the following salts

The question goes:

"What is the pH of a 0.1 M solution of each of the following salts? Before you begin, estimate the pH (neutral, acidic or basic).

a.MgF2
b.(NH4)2SO4
c.MgCl2
d.KCH3CO2"


If someone can get explain to me why they're basic, acidic, etc. as well, that would be very helpful! (that is, why they donate or receive protons; I know that acids give them and bases receive them)

MgF2 --> Mg 2+ + 2F-. F- will hydrolyze forming HF and OH-, HF is a weak acid, OH- is a strong base. basic pH

Ksp MgF2 = 7.4x10^-11 = [Mg 2+][F-]^2 = 4x^3
1.85x10^-11 = x^3
2.64x10^-4M = x
[Mg 2+] = 2.64x10^-4M, [F-] = 5.29x10^-4M

Ka HF = 6.3x10^-4 so the Kb of F- = 1.587x10^-11
Kb = [HF][OH-] / [F-]
1.587x10^-4 = x^2 / 5.29x10^-4 - x
x^2 + 8.4x10^-15x - 1.587x10^-4 = 0
x = [OH-] = 0.0126M
pH = 14 - pOH = 12.1


b. (NH4)2SO4 --> 2NH4+ + SO4 2-.....neutral
NH4+ forms NH3 + H+ in solution
SO4 2- will hydrolyze forming HSO4- and OH-
there are 2NH4+ for every 1 SO4 2- but SO4 2- has 2 Kb values


c. MgCl2 = acidic, Cl- comes from a strong acid, HCl, 0.1M MgCl2 = pH 1

d. KC2H3O2 = slightly basic
KC2H3O2 --> K+ + C2H3O2-
C2H3O2- will hydrolyze to form the weak acid, HC2H3O2 and OH-
Ka HC2H3O2 = 1.8x10^-5 so Kb = 5.56x10^-10
Kb = 5.56x10^-10 = [HC2H3O2][OH-] / [C2H3O2-]
5.56x10^-10 = x^2 / 0.1 - x
x^2 = 5.56x10^-12
x = [OH-] = 2.36x10^-5M
pH = 14 - pOH = 9.37