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Topic: phosphoric acid  (Read 2238 times)

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Offline Radu

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phosphoric acid
« on: February 25, 2014, 12:11:39 PM »
     I have a problem, I have to calculate the solubility ( the aparent solubility) of Ca3(PO4)2 in pure water, given ks=2.22*10-25.
      ka1=10-2.12493; ka2=10-7.2076; ka3=10-12.318758;
   
       Here's what I did:
                   Besides ionic dissociation of calcium phosphate, the following equilibrium establishes:
                           PO43-+H2O ::equil:: HPO42- + HO-. , K=Kw/ka3 = 10-1.6812 .

               Creating a table with the extent of this reaction at equilibrium, I obtain x as a function of aparent solubility, that is to say I solve the equation x2/2S-x = K.
             Then I write Ks= (3S)3*(2S-x(S))2, but it is not analitically solvable.
         Any easier ways of treating this problem?
             

Offline Sabinol

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Re: phosphoric acid
« Reply #1 on: March 31, 2014, 09:19:25 AM »
Shouldn't the solubility be:
2S = [HPO42-] + [PO43-]
and, knowing that K= Kw/Ka3 = [HPO42-][HO-]/[PO43-]
you will get :
2S = [PO43-] (1 + Kw/Ka3[HO-])
which is equivalent to:
2S = [PO43-] (1 + [H+]/Ka3)
and, because Ks = [Ca2+]3*[PO43-]2
substituting [PO43-] with something you can easily derive from the formula above and [Ca2+] = 3S, you will get :
Ks = (3S)3*(2S)2/(1 + [H+]/Ka3)2
which is a solvable equation if you know the solution's pH (in this case 7, obviously :)) )

Offline Radu

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Re: phosphoric acid
« Reply #2 on: March 31, 2014, 09:31:25 AM »
 It's not 7, not even near 7, that's the problem. There've been more discussions about this here:
            http://www.chemicalforums.com/index.php?topic=73202.0
   

Offline Sabinol

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Re: phosphoric acid
« Reply #3 on: March 31, 2014, 09:39:39 AM »
Yes, I was just looking on the topic.. Seems like I underestimated the problem

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