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Author Topic: Precipitation problem  (Read 724 times)

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moop1

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Precipitation problem
« on: December 18, 2016, 08:06:55 AM »

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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AWK

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Re: Precipitation problem
« Reply #1 on: December 18, 2016, 08:25:56 AM »

Calculate stoichiometry of percipitation reaction and excessing concentration of Na3PO4. Use this concentration in solubility product expression.
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moop1

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Re: Precipitation problem
« Reply #2 on: December 18, 2016, 08:34:11 AM »

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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AWK

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Re: Precipitation problem
« Reply #3 on: December 18, 2016, 10:09:22 AM »

These are concentrations before reaction.
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moop1

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Re: Precipitation problem
« Reply #4 on: December 18, 2016, 11:42:43 AM »

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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AWK

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Re: Precipitation problem
« Reply #5 on: December 18, 2016, 12:41:46 PM »

Show your calculations. Stoichiometry of precipitation reaction is very important!
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Borek

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Re: Precipitation problem
« Reply #6 on: December 18, 2016, 12:50:08 PM »

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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moop1

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Re: Precipitation problem
« Reply #7 on: December 19, 2016, 04:56:02 AM »

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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AWK

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Re: Precipitation problem
« Reply #8 on: December 19, 2016, 05:05:38 AM »

???
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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moop1

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Re: Precipitation problem
« Reply #9 on: December 19, 2016, 05:23:29 AM »

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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AWK

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Re: Precipitation problem
« Reply #10 on: December 19, 2016, 05:36:15 AM »

Ca2+ is removed completely. Check you stoichiometry calculations
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moop1

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Re: Precipitation problem
« Reply #11 on: December 19, 2016, 05:38:01 AM »

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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AWK

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Re: Precipitation problem
« Reply #12 on: December 19, 2016, 05:44:23 AM »

3Ca2+ = 1.80*10-3
2Ca2+ is an equivalent of removed phosphate
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moop1

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Re: Precipitation problem
« Reply #13 on: December 19, 2016, 06:39:23 AM »

So what are you suggesting ? That I should divide by 3 ? It also gives me a negative.
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AWK

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Re: Precipitation problem
« Reply #14 on: December 19, 2016, 06:45:27 AM »

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
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