Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek 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?

Anybody there?

Come on guys, it is not THAT difficult. Half of the information is there just to muddy the water ;)

is it 62.93ml?

Sorry, no.
But it would be interesting to learn how you got this particular volume?

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
[tex]pH=pKa+log\frac{[CH_3COO^]}{[CH_3COOH]}=4.71[/tex]
[tex][H^+]_{initial}=1.9498*10^{5}[/tex]
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 [tex]\frac{[CH_3COO^]}{[CH_3COOH]}=0.17378[/tex]
based on the Ka for CH_{3}COOH,
[tex]1.7378*10^{5}=\frac{[CH_3COO^][H^+]}{[CH_3COOH]}[/tex]
assume that x mol of HCl was added, then after that y mol of the CH_{3}COOH reformed to achieve new equilibrium,
[tex]1.7378*10^{5}=\frac{([H^+]+xy)([CH_3COO^]y)}{[CH_3COOH]+y}[/tex]
[tex]\frac{[CH_3COO^]y}{[CH_3COOH]+y}=0.17378[/tex]
so [itex][H^+]+xy=1*10^{4}[/itex]
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
[itex]\frac{[CH_3COO^]}{[CH_3COOH]}[/itex] does not depend on the volume of solution, all the NaOH (which is 0.10878mol) and 0.10878+8.0502x10^{5}mol 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!

You felt into a trap.
Hint: these are way too concentrated solutions to be used directly.

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.

my guess is around 1387.6 L
And that's the answer I was looking for :)

yay!
i hope its not cheating that im a second year and this forum says its for posting first year questions :P

Borek what happened to Problem of the week?

Not enough interest to continue. This particular problem required a month, not a week.
We may start again one day.

I was busy that time, but I would like to attempt now any PoW you post.

This particular question is somewhat complicated for me but from most of the past PoW questions I would be happy to attempt more.