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In: Chemistry

Calculate the solubility of Mn(OH)2 in grams per liter when buffered at pH=7.0, 9.8, 11.8. Assume...

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.

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Expert Solution

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


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