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Topic: Back with the phosphoric acid- problem 11  (Read 6901 times)

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Offline Radu

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Back with the phosphoric acid- problem 11
« on: March 31, 2014, 08:51:27 AM »
http://www.chemicalforums.com/index.php?topic=73202.0

   Here's the old topic, from what I've read everything is pretty clear except 4.
   
      I just want to know if my assumptions are correct(maybe you have used an equation solver to get the accurate answer, I didn't find one to  deal with the explicit form of the equation, maybe you can recommend me one? ):
                        Ksp=[Ca2+]3[PO43-]2

       Now, a small amount of phosphate dissociates: PO43- + H2::equil::  HPO42- + HO-. Assuming that [OH-]=[ HPO42-], I have x2/(S-x)= K=kw/ka3=10-1.6812.
     Therefore S= (x2+ Kx)/K. So, [Ca2+]=S= (x2+ Kx)/K and [PO43-]=S-x=x2/K.
       Everything right until here, but here comes the interesting part: Given we have a saturated solution of a poorly soluble compound, we expect the pH to be less than , say, 10 ( the idea is that [HO-]<<K=10-1.6812, and therefore S≈x). Hence, we obtain a solvable equation:
         x7=Ksp*K2, which leads to x=10-4.0023, and S=9.9946*10-5.
           I also obtain pH=9.9977, which is in accordance with my assumption( of course, mathematically, this doesn't prove anything, but I've seen that most problems deal with their assumptions just like this)
       What do you think?
         
       

         

Offline Rutherford

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Re: Back with the phosphoric acid- problem 11
« Reply #1 on: March 31, 2014, 10:51:34 AM »
S≈x meaning that PO43- almost completely hydrolyzed, is a good assumption. I think that the phosphate can be regarded as a strong base, and since it's concentration is low, the assumption is valid. This seems wrong:
"Now, a small amount of phosphate dissociates:"
Actually almost the whole amount of the phosphate dissociates, but the concentration will still be small.
And if [Ca2+]=S then this can't be true [PO43-]=S-x, becuse when one unit of the salt dissociates three cations and two anions are made, not equal amount of both.
I like your idea. They probably don't expect us to do the algebra, but rather to think of a good assumption.

Offline Rutherford

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Re: Back with the phosphoric acid- problem 11
« Reply #2 on: March 31, 2014, 11:10:13 AM »
I did the calculation with corrected coefficients and got that S=1.14·10-4M which is close to 1.26·10-4M which I got using wolfram. Close enough for me.
So it's actually: S=3/2x
« Last Edit: March 31, 2014, 11:58:55 AM by Raderford »

Offline Radu

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Re: Back with the phosphoric acid- problem 11
« Reply #3 on: March 31, 2014, 12:15:51 PM »
    How did you get your results?
           Because, working with the coefficients: K=x2/(2S-x), whence S=(x2+Kx)/2K.
     Given Kx>>x2, S=x/2, meaning all phosphate hydrolised. Now the concentration of Ca2+ is 3S=3x/2., and [PO43-]=2S-x=x2/K.
          Inserting, I get ksp=27/8*x7*1/K2. x=8.3604*10-5
, and S=4.1802*10-5
        And yes, when I said that a small part of phosphate dissociates I was talking in absolute terms, not relative to x.

Offline Rutherford

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Re: Back with the phosphoric acid- problem 11
« Reply #4 on: March 31, 2014, 12:32:44 PM »
[PO42-]=2/3S-x
K=x2/(2/3S-x) and S=1.5·x(x+K)/K
Using the approximation:
S=1.5x
[PO42-]=2/3S-x=x2/K
Then:
Ksp=27/8·x3·x4/K2
And I got that x=8.38·10-5M, meaning that S=1.26·10-4M
I had a little mistake before. Now it is absolutely the same answer I got with a program that operates with the complicated algebra.

Offline Radu

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Re: Back with the phosphoric acid- problem 11
« Reply #5 on: March 31, 2014, 12:46:29 PM »
[PO42-]=2/3S-x
K=x2/(2/3S-x) and S=1.5·x(x+K)/K
Using the approximation:
S=1.5x


 Why [PO42-]=2/3S-x , when the total concentration of phosphate is 2S and the amount consumed in the reaction is simply x? so we'd get: [PO43-]=2S-x, and S≈x/2.

Offline Rutherford

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Re: Back with the phosphoric acid- problem 11
« Reply #6 on: March 31, 2014, 01:00:05 PM »
If S=[Ca2+] and per three calcium ions two phosphate ions are produced then the starting concentrations of phosphate will be 2/3S.

Offline Radu

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Re: Back with the phosphoric acid- problem 11
« Reply #7 on: March 31, 2014, 01:13:48 PM »
 Ah, I totally agree with you,my result is the moles of calcium phosphate that dissociates, and your result is simply three times greater than mine. But I think the solubility refers to the number of moles of salt that dissociates,  why would you refer to the free calcium in solution?

Offline Rutherford

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Re: Back with the phosphoric acid- problem 11
« Reply #8 on: March 31, 2014, 01:29:10 PM »
Sorry, I got confused there.

Offline Rutherford

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Re: Back with the phosphoric acid- problem 11
« Reply #9 on: May 05, 2014, 10:19:44 AM »
Is this assumption valid:
[HPO42-]>>[PO43-]
EDIT: Now I think not, the relative difference in concentrations of these two isn't so big.
« Last Edit: May 05, 2014, 12:34:43 PM by Raderford »

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