In: Chemistry
Calculate the solubility of Mn(OH)2 in grams per liter when buffered at pH=7.0, 9.8, 11.8. Assume that buffer capacity is not exhausted.
at
pH = 7.0
pOH = 14 - pH = 7.0
pOH = -log[OH-]
7.0 = -log[OH-]
[OH-] = 1.0 x 10^-7 M
Ksp of Mn(OH)2 = 2 x 10^-13
Mn(OH)2 ------------------> Mn+2 + 2OH-
Ksp = [Mn+2][OH-]^2
1.6 x 10^-13 = [S][1.0 x 10^-7]^2
S = 16 mol/liter
molecular mass of Mn(OH)2 = 88.95 g/mol
solubility S = 16 x 88.95 = 1423.2 g/L
2) pH = 9.8
pOH = 14 - 9.8 = 4.2
[OH-] = 6.31 x 10^-5 M
Ksp = [Mn+2][OH-]^2
1.6 x 10^-13 = S x (6.31 x 10^-5)^2
1.6 x 10^-13 = S x (3.98 x 10^-9
S = 4.02 x 10^-5 M
solubity S = 88.95 x 4.02 x 10^-5
= 3.56 x 10^-3 g/L
c) pH = 11.8
pOH = 14 - 11.8 = 2.2
[OH-] = 6.31 x 10^-3 M
Ksp = [Mn+2][OH-]^2
1.6 x 10^-13 = S x (6.31 x 10^-3)^2
1.6 x 10^-13 = S x (3.98 x 10^-5)
S = 4.02 x 10^-9 M
solubity S = 88.95 x 4.02 x 10^-9
= 3.56 x 10^-7 g/L