Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: stefufufu on October 26, 2013, 10:48:22 AM
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Ca3(PO4)2 has a solubility product Ksp = 1.0x10-25. Calculate the molar concentration of calcium ions in solution in equilibrium with solid Ca3(PO4)2.
Ca3(PO4)2 Ksp = 1.0x10-25
Ca3(PO4)2 ::equil:: 3Ca+ + 2PO43-
Ksp = [Ca2+]3[PO43-]2 = 1.0x10-25
= [3x]3[2x]2 = 108x5 = 1.0x10-25
[Ca2+] = 3x = 3 x 5√1.0x10^-25/108 = 1.176x10-5 moles/litre.
This is as far as I have managed to use notes to work moles/litre out. I'm not sure which way to work out the Molar concentration of the calcium, because I do not have a volume? M = n/V.
I feel I'm going completely off track.
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in your opinion, is there any difference at all between the " moles/litre " you've calculated for [Ca2+] ( and correctly at that, insofar as the systematic forced on you requires it) and the molar concentration M ?
regards
Ingo
additional remark I: the systematic forced on you in this calculation leads to utterly, utterly wrong results, mostly because PO43- won't remain what it initially was, but would become protonated next to completely.
to a minor degree, this also is true for Ca2+ , as this becomes hydroxylated
additional remark II : the Ksp given to you is wrong by several orders of magnitude.
the original value in literature is 2,07 * 10-33 moles5/L5 at 25 °C
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x is the molar concentration and is in moles per litre
Ksp is the equilibrium constant and the expression has quantities in molar concentrations
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x is the molar concentration and is in moles per litre
yes , it is
and 3 x is exactly the concentration of Ca2+ , resulting of x moles of Ca3(PO4)2
hence:
[Ca2+] = 3x = 3 x 5√1.0x10^-25/108 = 1.176x10-5 moles/litre.
this already is the concentration you're asked to calculate for:
(http://Calculate the molar concentration of calcium ions in solution)
you're already there
regards
Ingo
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you're already there
Who? Vidya is not the OP.
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hence:
[Ca2+] = 3x = 3 x 5√1.0x10^-25/108 = 1.176x10-5 moles/litre.
this already is the concentration you're asked to calculate for:
(http://Calculate the molar concentration of calcium ions in solution)
you're already there
regards
Ingo
So, I have my answer already? Great!
Thank you all for your replies.