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Topic: calculating solubility in mol/L of Calcium Phosphate in water  (Read 9105 times)

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Offline NewtoAtoms

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calculating solubility in mol/L of Calcium Phosphate in water
« on: February 03, 2009, 10:03:55 PM »
Hello Chemists,

I once again have finished Chapter 16, solubility equilibria chapter in my text, and have stumbled across a perplexing problem, and hoping someone can help me walk through. Unfortunately my chemistry teacher doesn't speak English so us students are left to our own devices to learn this stuff. 

Q:  Using the Ksp values given, find the solubility in mol/L of Calcium Phosphate in water.

A:  I have done the following work, and I am wondering if someone can review it with me and aid me on the issues?

Ca3(PO4)2 <----------> 3Ca2+  +  2(PO4)3-

Ksp = [Ca]3 [PO4]2

                     Ca                       PO4
Initial               O                         O
Change            +3s                       +2s
Equilibrium        3s                          2s

Ksp = [3s]3 [2s]2 = (27s3)(4s2) = 108 s5

Ksp for this particular reaction is 1.2 x 10-26 therefore

1.2 x 10-26 = 108 s5 (divide both sides by 108)
1.11 x 10-28 = s5  (fifth root each side)
gives the answer. 

The problem is that I am not even sure if I have done this correct, before I try to venture out and try to learn if there is actually something called a fifth root.

I would greatly appreciate anyone's help and guidance.

Thank you!

Offline AWK

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Re: calculating solubility in mol/L of Calcium Phosphate in water
« Reply #1 on: February 04, 2009, 02:07:55 AM »
OK
AWK

Offline NewtoAtoms

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Re: calculating solubility in mol/L of Calcium Phosphate in water
« Reply #2 on: February 04, 2009, 05:31:18 PM »
OK?  Does that mean I have done this correctly and am understanding?

Thank you for your time.

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