Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: moop1 on December 18, 2016, 02:06:55 PM
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We have 5 L of water that contains Ca2+=1,80*10-3 mol/L in which we put 8,23 g of Na3PO4. We see the fromation of a precipitate.
Find the concentration of Ca2+ after precipitation.
I know that the precipitate is : Ca3(PO4)2 with Kps = 2,07*10-33
Besides that, I'm stuck. ANy hints?
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Calculate stoichiometry of percipitation reaction and excessing concentration of Na3PO4. Use this concentration in solubility product expression.
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Yeah but what are the concentrations ?
I found this 1,80 *10^-3 for Ca
and 0,01004 for PO4
Is there something wrong here ?
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These are concentrations before reaction.
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But are they the good ones ? I know that the answer to the problem is 6,35*10^-9 but I keep getting 6*10^-4 . Clearly Im doing the calculation with the wrong concentraitons.
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Show your calculations. Stoichiometry of precipitation reaction is very important!
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But are they the good ones ?
No, because they don't take the precipitation into account.
AWK told you twice to use the precipitation stoichiometry to describe changes taking place in the solution. As long as you ignore this advice you will not move forward.
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You mean you want to multiply by the number of composition ?
Here's my stoichio graph :
Ca3(PO4)2 = 3Ca2+ + 2 PO4
Ca(PO4)2 = 0 M
the Ca and PO4 Im not sure how to obtain them. We know that we're supposed to obtain a rest of Ca. so I think I will most likely have no PO4 after the reaction happened
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???
Reaction of synthesis is:
Ca2+ + PO43- = Ca3(PO4)2 (unbalanced)
In your case this is reaction with limited reagent. Forget solubility equilibrium at the moment and calculate concentration of excessing reagent after reaction as if the reaction went to completion.
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So molar mass of Na3PO4 = 163,94 g/mol
8,23 g * 1 mol / 163,94 g = 0,050201 mol
after I put it in the same unit as the other concentration : 0,050201 mol /5 L = 0,01004 mol/L
after
Ca3(PO4)2 = 3Ca2+ + 2 PO4 3-
0 1,80*10^-3 0,01004
+0,01004 - 5*10^-3 -0,01004
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Ca2+ is removed completely. Check you stoichiometry calculations
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Yeah, I know. But that's where I have been to be able to go alone, after that I get stuck. I thought dividing by 2 and substracting would be enough but ti doesn't seem so. :/
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3Ca2+ = 1.80*10-3
2Ca2+ is an equivalent of removed phosphate
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So what are you suggesting ? That I should divide by 3 ? It also gives me a negative.
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These are very elementary stoichiometric operations. You should go back to textbook and read chapter on stoichiometry of reaction.
A good stsrt is here:
http://www.chembuddy.com/?left=balancing-stoichiometry&right=toc