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Topic: Solubility Question  (Read 3927 times)

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Offline Ayalon

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Solubility Question
« on: March 12, 2010, 12:58:59 PM »
I'd really appreciate it if someone could help me solve this one:

What would the concentration of Phosphate Ions be in a mixture of 1.5 mol Mg3(PO4)2 in water with volume 2.0L? (Ksp=1.04x10^-24)

Thanks a lot!

Offline Borek

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Re: Solubility Question
« Reply #1 on: March 12, 2010, 02:50:57 PM »
Ksp formula?

How is concentration of PO43- related to concentration of Mg2+?

Are you expected to neglect hydrolysis, or not?
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Offline Ayalon

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Re: Solubility Question
« Reply #2 on: March 12, 2010, 03:07:47 PM »
Mg3(PO4)2 ::equil:: 3Mg^2+ & 2PO4^3-

Not sure how I'm supposed to get one concentration from the other, also not sure if hydrolysis can be neglected.

Offline Borek

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Re: Solubility Question
« Reply #3 on: March 12, 2010, 03:19:14 PM »
Mg3(PO4)2 ::equil:: 3Mg^2+ & 2PO4^3-

Not sure how I'm supposed to get one concentration from the other

Stoichiometry. Take a look at the reaction equation.

Write Ksp formula.

Quote
also not sure if hydrolysis can be neglected.

Can't decide for you.
« Last Edit: March 12, 2010, 04:50:17 PM by Borek »
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Offline Ayalon

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Re: Solubility Question
« Reply #4 on: March 12, 2010, 03:28:38 PM »
(3S)^3*(2S)^2= 108S^5 = 1.04x10^-24
S = 6.262x10^-6 M
Don't know how to reach 2PO4 concentration based on Mg3(PO4)2 concentration...

Offline Borek

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Re: Solubility Question
« Reply #5 on: March 12, 2010, 04:53:37 PM »
AMount of solid in mixture doesn't matter - that is, if there is enough of the solid to make saturated solution, it won't get "more" saturated if you add more solid.

As I read the question you have a suspension prepared by agitating solid magnesium pohsphate (1.5 moles) with water (2L), If so, your approach looks OK. Just note that so far you have calculated solubility of the salt - that's not the same as concentration of phosphate. Again, take a look at stoichiometry.
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