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### Topic: Problem of the week - 03/06/2013  (Read 28492 times)

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#### Borek

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##### Problem of the week - 03/06/2013
« on: June 03, 2013, 03:48:13 PM »
You are given 10 mL of a glacial acetic acid (density 1.0497 g/mL), 40 mL of a 28% M sodium acetate (density 1.1462 g/mL), 10 mL of a 35% hydrochloric acid (density 1.174 g/mL), 6 mL of a 48% sodium hydroxide solution (density 1.5109 g/mL), and asked to prepare as much pH=4.00 solution as possible.

What is the maximum volume of the solution?
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#### Borek

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##### Re: Problem of the week - 03/06/2013
« Reply #1 on: June 17, 2013, 05:02:58 AM »
Anybody there?
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#### Borek

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##### Re: Problem of the week - 03/06/2013
« Reply #2 on: June 24, 2013, 05:03:44 AM »
Come on guys, it is not THAT difficult. Half of the information is there just to muddy the water
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#### SinkingTako

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##### Re: Problem of the week - 03/06/2013
« Reply #3 on: June 27, 2013, 04:15:55 AM »
is it 62.93ml?
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#### Borek

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##### Re: Problem of the week - 03/06/2013
« Reply #4 on: June 27, 2013, 04:36:01 AM »
Sorry, no.

But it would be interesting to learn how you got this particular volume?
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#### SinkingTako

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##### Re: Problem of the week - 03/06/2013
« Reply #5 on: June 27, 2013, 10:44:26 PM »
Now I get another answer that seems to be 65.67ml...

Firstly I found the amount of each chemical present:

0.17480mol of Acetic acid
0.15670mol of sodium acetate
0.11258mol of HCl
0.10878mol of NaOH

pKa of Acetic acid is 4.76
assume that all of the acetic acid and sodium acetate were used, the pH of the solution will be
$$pH=pKa+log\frac{[CH_3COO^-]}{[CH_3COOH]}=4.71$$
$$[H^+]_{initial}=1.9498*10^{-5}$$
Thus more acid will need to be added. Assume that the pH of the solution is 4.00, and that HCl and NaOH are strong acid/bases and don't contribute to the buffer capacity
then $$\frac{[CH_3COO^-]}{[CH_3COOH]}=0.17378$$

based on the Ka for CH3COOH,
$$1.7378*10^{-5}=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$$
assume that x mol of HCl was added, then after that y mol of the CH3COOH reformed to achieve new equilibrium,

$$1.7378*10^{-5}=\frac{([H^+]+x-y)([CH_3COO^-]-y)}{[CH_3COOH]+y}$$

$$\frac{[CH_3COO^-]-y}{[CH_3COOH]+y}=0.17378$$
so $[H^+]+x-y=1*10^{-4}$
I assumed that x<<y, not too sure whether it's a valid assumption
so solving, x=8.0502x10-5
so this amount of HCl was added.

Since NaOH and HCl are strong and addition in equal molarity only adds to the water content, and
$\frac{[CH_3COO^-]}{[CH_3COOH]}$ does not depend on the volume of solution, all the NaOH (which is 0.10878mol) and 0.10878+8.0502x10-5mol of HCl can be added, which is 9.67ml

so total volume of buffer formed is 10+40+9.67+6=65.67ml

I think there's something fishy/wrong assumptions/calcuations made though. Thanks!
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#### Borek

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##### Re: Problem of the week - 03/06/2013
« Reply #6 on: June 28, 2013, 03:17:30 AM »
You felt into a trap.

Hint: these are way too concentrated solutions to be used directly.
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#### Squeegy

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##### Re: Problem of the week - 03/06/2013
« Reply #7 on: June 30, 2013, 08:06:14 AM »
so i think i have it, but i could be wrong, so ill explain the process i went through.
1) worked out the number of moles of HCl, which was about .113
2) worked out the number of moles of glacial acetic acid, which was about .175
3) worked out the number of moles of H+ that would dissociate from the acetic acid at pH 4, using hasselbalch equation, got about .0259
4) used the total mole of H+ in the pH equation, replacing [H+] with (mole H+/volume) and a pH of 4, and solving for volume, giving 1387.6 L

so my guess is around 1387.6 L (i didnt used rough atomic weights since i didnt have a periodic table handy, and may have messed up other math)

in practical terms, you just put both the acids in a BIG container and dilute till it has a pH of 4.

#### Borek

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##### Re: Problem of the week - 03/06/2013
« Reply #8 on: July 01, 2013, 10:20:01 AM »
my guess is around 1387.6 L

And that's the answer I was looking for
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#### Squeegy

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##### Re: Problem of the week - 03/06/2013
« Reply #9 on: July 03, 2013, 10:46:38 AM »
yay!
i hope its not cheating that im a second year and this forum says its for posting first year questions

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##### Re: Problem of the week - 03/06/2013
« Reply #10 on: September 21, 2013, 11:19:26 AM »
Borek what happened to Problem of the week?

#### Borek

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##### Re: Problem of the week - 03/06/2013
« Reply #11 on: September 23, 2013, 12:12:05 PM »
Not enough interest to continue. This particular problem required a month, not a week.

We may start again one day.
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#### Rutherford

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##### Re: Problem of the week - 03/06/2013
« Reply #12 on: September 23, 2013, 12:59:10 PM »
I was busy that time, but I would like to attempt now any PoW you post.