Solubility Problem?

Question

Calculate the solubibilty of Mn(OH)2 in grams per liter when buffered at pH a) 7.0, b)9.5, c)11.8


Mn(OH)2 Ksp=1.6x10^-13

I thought I was doing it right , but my numbers are messed up.

Answer 1

Actually you shouldn't solve quadratics.
You have a buffered system, thus the concentration of OH- at equilibrium is not going to be x+10^-pOH, but simply 10^-pOH; the buffer keeps the pH value and therefore the [OH-] constant.

The solubility (x) is going to be equal to the concentration of Mn+2

Mn(OH)2 <=> Mn+2 + 2OH-

Ksp=[Mn+2][OH-]^2 =>
solubility x = [Mn+2]= Ksp/[OH-]^2 =Ksp/ (10^-pOH)^2 =
= Ksp/10^-2pOH= Ksp /(10^-2*(14-pH))

a) x =(1.6*10^-13)/(10^-(2*(14-7))) = 16 M
b) x =(1.6*10^-13)/(10^-(2*(14-9.5))) = 1.6*10^-4 M
c) x =(1.6*10^-13)/(10^-(2*(14-11.8))) =4.0* 10^-9 M

Answer 2

OK. you have Ksp = [Mn] [OH]^2 = 1.6x10^-13
Let's say x=solubility of Mn(OH)2

for part a, you solve for x
x(2x)^2 = 1.6x10^-13
You get x=3.4 x 10^-5 mol/L

B) [OH] added is 10^-4.5= 3.16 x 10^-5 M
you solve for x
x(2x + 3.16x10^-5)^2=1.6x10^-13
x = 2.46 x 10^-5

C) [OH} added =10^-2.2 = 0.0063 M
x(2x + 0.0063)^2=1.6x10^-13
x=4.03x10-9

The quadratics that I did above was solved using a graphic solver