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Offline Nagenyleveable

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Concentrations
« on: August 20, 2010, 10:42:53 AM »
One litre of aqueous sample containing a specific organic pollutant B was brought to a lab for analysis. two extractions each with 50ml of fresh organic solvent were performed. on analysis of the organic solvent after the first extraction the concentration of B was 6µg/ml. after the second extraction with fresh solvent, the concentration of B was 0.5µg/ml

(i) Determine the concentration of B in the original aqueous sample solution.

(ii) Determine the distribution ratio of B in the organic solvent and aqueous solution.

(iii) Determine the minimum number of extractions needed to achieve an extraction efficiency of greater than 99%(each performed with 50ml of fresh organic solvent).


Can anyone help with the formulas and break down of the figures for this question please, im really stuck!
Thanks

Offline Nagenyleveable

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Re: Concentrations
« Reply #1 on: August 20, 2010, 03:27:35 PM »
OK so far i know the equation for part two D=[(wo-w1)/y]/w1/x

with x=1 y=0.05 and wo=? and w1=?

 and for part three its going to be
(w0-wn)/w0>99%=

wn/wo= (x/Dy+x)^n<0.02

So its just the first part im confused about and the values of wo and w1 (which im assuming are the concentrations)

Offline MrTeo

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Re: Concentrations
« Reply #2 on: August 21, 2010, 02:29:43 AM »
OK so far i know the equation for part two D=[(wo-w1)/y]/w1/x

with x=1 y=0.05 and wo=? and w1=?

If you want to solve an equation with two unknown variables (D and w0) you need to set up a system, the equation you wrote it's OK (I don't really know if you inverted the volumes or if I just misread your expression, anyway here it is, this should be the correct form):



Now write the equation of the second extraction phase and try to solve the system of these two: you'll easily find both the distribution ratio (D) and the starting concentration, then we'll think about the third point  ;)
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Offline Nagenyleveable

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Re: Concentrations
« Reply #3 on: August 21, 2010, 09:49:32 AM »
I dont follow this. If I was able to work out the concentrations in part i (which is my biggest concern) would this not give me the values of w0 and w1 for the second part?

Offline MrTeo

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Re: Concentrations
« Reply #4 on: August 21, 2010, 09:53:13 AM »
...

If you have found out w0 and w1 the first and the second part are done...

So its just the first part im confused about

Could you please explain what did you found and what do you need?  ;D
The way of the superior man may be compared to what takes place in traveling, when to go to a distance we must first traverse the space that is near, and in ascending a height, when we must begin from the lower ground. (Confucius)

Offline Nagenyleveable

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Re: Concentrations
« Reply #5 on: August 21, 2010, 10:08:21 AM »
ok I dont know how to do part (i) which is to work out the concentration of B in the original sample.

I am assuming that for the second part(ii) the equation I would use is:
w1= w0(x/Dy+x)   with x=1 y=0.05 but im not really sure what w1 and w0 represent

the third part im pretty confident about but again i need to know what w1 and w0 represent to be sure. In the sample question that im working from they seem to represent the mass of the organic compound which im assuming i could work out if i know the concentrations?

Offline MrTeo

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Re: Concentrations
« Reply #6 on: August 22, 2010, 01:08:58 PM »
OK so far i know the equation for part two D=[(wo-w1)/y]/w1/x

with x=1 y=0.05 and wo=? and w1=?

To sum up: this is ok, though seems to me you've inverted the volumes: x should be 0.05 L and y 1 L, I think...

Now, w0 and w1 are the concentrations (the first is the starting one, the second tells us the amount of compund extracted and it's 6µg/L)
If you do the same thing only with the second extraction (which gives us a w2 of 0.5µg/L) you'll find another equation and you'll be able to set up a two-equation system to find out D and w0, the answers for the first two points of the exercise.

Try this and then we'll think about the third one: you can't do the third without having understood the other two.
The way of the superior man may be compared to what takes place in traveling, when to go to a distance we must first traverse the space that is near, and in ascending a height, when we must begin from the lower ground. (Confucius)

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