Can someone help me with this?

Calculate the solubility of Mn(OH)2 in grams per liter when buffered at a pH (a) 7 (b) 10

Answer:
Ksp of Mn(OH)2 = 1.9 x 10^-13
the equilibrium is
Mn(OH)2 <> Mn2+ + 2OH-
and the requirement for this equilibrium is
Ksp = [Mn2+][OH-]^2
at pH = 7 we have [OH-] = 1 x 10^-7 M
1.9 x 10^-13 = [Mn2+] ( 1 x 10^-7)^2
The base is soluble

at pH = 10 we have [OH-] = 1 x 10^-4 M
1.9 x 10^-13 = [Mn2+] ( 1 x 10^-4)^2
[Mn2+] = 1.9 x 10^-5 M
no, I will not help you with this

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