Forming precipitate in buffer solution

[thatgirlyouknow]

Homework Statement

The solubility of a hydroxide M(OH)2 is .0090 g/liter. Molecular weight of M(OH)2 is 150. If 1.0 g of MSO4 is dissolved in 500 ml of water buffered at pH 8.5, will a precipitate form?

Homework Equations

pOH + pH = 14
molarity = mol/liter

The Attempt at a Solution

pOH must be 5.5, so the OH concentration is 10^-5.5 = 3.1623x10^-6.
Solubility of M(OH)2 is .00006 mol/liter. I don't know where the buffer solution fits into all of this and how to know if a precipitate forms or not.
MSO4 <--> M + SO4
s s s
Ksp = s^2
How can I find the solubility of MSO4? Is that even necessary?

 

[chemisttree]

Assume that MSO4 is completely soluble at a concentration of 1 g/500 mL, unbuffered.

Write out the equilibrium expression of M(OH)2 and determine the Ksp of the M(OH)2 using the data you are given. Calculate the [OH-] at pH 8.5 and substitute it into the Ksp expression and see if the Ksp has been exceeded.

 

[thatgirlyouknow]

M(OH)2 <--> M + 2OH
.0090 g/liter * 1mol/150 g = 6e-5
substituting this in, we get
4s^3 = Ksp = 8.64e-13
pOH = 5.5 = 10^-5.5 = 3.16228 concentration of OH-
So then where does this go? Do I put it into
[M][3.1623e-6]^2 = Ksp
What then do I use for the concentration of M? The original solubility? This gives:
[6e-5][3.1623e-6]^2 = 6e-16

Then where does the SO4 come into play? Its concentration must be important too, otherwise I wouldn't have been given the 1 gram and 500 mL.

 

[chemisttree]

M(OH)2 <--> M + 2OH
.0090 g/liter * 1mol/150 g = 6e-5
substituting this in, we get
4s^3 = Ksp = 8.64e-13
pOH = 5.5 = 10^-5.5 = 3.16228 concentration of OH-
So then where does this go? Do I put it into
[M][3.1623e-6]^2 = Ksp
What then do I use for the concentration of M? The original solubility? This gives:
[6e-5][3.1623e-6]^2 = 6e-16

Then where does the SO4 come into play? Its concentration must be important too, otherwise I wouldn't have been given the 1 gram and 500 mL.

Given that the molecular weight of M(OH)2 is 150 (and OH is 17.01), can you determine the atomic mass of M? That is where the sulfate comes into play. Determine the molecular weight of MSO4. You know the mass (1g) and can calculate the [M+2] from that.

The formula for Ksp in this case is

Ksp = [M+2][OH-]^2

If only 0.0090 grams per liter is soluble and the formula weight is 150, then the number of moles of M(OH)2 (and M+2) will indeed be 6e-5. How many moles of OH- will there be?

Your equation will tell you that.

M(OH)2 <-----> M+2 + 2OH-

Substitute that value into the equation for Ksp and you will get the Ksp value. It's not 8.64e-13.

Use the concentration of M+2 given when 1 g of MSO4 dissolves in 500 mL and assume that all of the OH- comes from the pOH as you have indicated. Remember to use the formula weight of MSO4 that you derive.