Forming precipitate in buffer solution

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In summary: Given that there are 3.16228 moles of OH- and the molarity of M(OH)2 is .00006 mol/liter, the concentration of MSO4 will be 3.16228 * 0.00006 = 0.00024 mol/liter. In summary, the hydroxide M(OH)2 is slightly soluble in water and will form a precipitate if dissolved in a solution with a high OH concentration. The molecular weight of M(OH)2 is 150 and the solubility of MSO4 is 0.00024 mol/liter.
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thatgirlyouknow
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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?
 
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  • #2
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.
 
  • #3
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.
 
  • #4
thatgirlyouknow said:
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.
 

FAQ: Forming precipitate in buffer solution

1. How does precipitate form in a buffer solution?

Precipitate forms in a buffer solution when the concentration of a solute exceeds its solubility in the solution. The excess solute molecules then come together and form solid particles called precipitates.

2. Can precipitate formation be prevented in a buffer solution?

Precipitate formation in a buffer solution can be prevented by maintaining the solute concentration below its solubility limit, using a different buffer system with a different pH, or adding chelating agents or surfactants to the solution.

3. What factors affect the formation of precipitate in a buffer solution?

The concentration of the solute, the pH of the buffer solution, and the presence of other ions or molecules that can interact with the solute can all affect the formation of precipitate in a buffer solution.

4. How can I determine if a precipitate has formed in a buffer solution?

You can determine if a precipitate has formed in a buffer solution by visually inspecting the solution for any changes in clarity or color. You can also perform a simple filtration or centrifugation test to separate the precipitate from the solution.

5. Is precipitate formation reversible in a buffer solution?

Precipitate formation in a buffer solution can be reversible if the conditions that caused the formation of precipitate are changed. For example, if the pH of the solution is adjusted or the concentration of the solute is reduced, the precipitate may dissolve back into the solution.

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