Question

Consider an amphoteric hydroxide, M(OH)2(s), where M is a generic metal. M(OH)2(s) <--> M2+(aq) + 2OH-(aq)...

Consider an amphoteric hydroxide, M(OH)2(s), where M is a generic metal.

M(OH)2(s) <--> M2+(aq) + 2OH-(aq) Ksp= 3x10-16

M(OH)2(s) +2OH-(aq) <--> [M(OH)4]2-(aq) Kf= 0.05

Estimate the solubility of M(OH)2 in a solution buffered at pH = 7.0, 10.0, and 14.0.

Homework Answers

Answer #1

the solubility for Mg(OH)2 where the pH is l0

Ksp = 3 x l0^-16 = (Mg++)(OH)^2 Mg(OH)2(s) at equilib Mg++(aq) + 2 OH-(aq)

If the pH is l0 then the pOH is 4 because pH + pOH = 14

and if the pOH is 4, then the (OH-) concentration = antilog(-pOH)

antilog (-4) = l x l0^-4

Ksp = 3 x l0^-16 = (Mg++)(l x l0^-4)^2

(Mg++) = 3 x l0^-16 / 1 x l0^-8 = 3 x l0^-8 Moles per liter solubility
Mg(OH)2 when pH is l0

At pH =7

the solubility for Mg(OH)2 where the pH is 7

Ksp = 3 x l0^-16 = (Mg++)(OH)^2 Mg(OH)2(s) at equilib Mg++(aq) + 2 OH-(aq)

If the pH is 7 then the pOH is 4 because pH + pOH = 14

and if the pOH is 7, then the (OH-) concentration = antilog(-pOH)

antilog (-7) = l x l0^-7

Ksp = 3 x l0^-16 = (Mg++)(l x l0^-7)^2

(Mg++) = 3 x l0^-16 / 1 x l0^-8 = 3 x l0^-14 Moles per liter solubility
Mg(OH)2 when pH is 7

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