Tuesday
July 29, 2014

Homework Help: science - chemistry

Posted by ajones51 on Sunday, October 8, 2006 at 12:52pm.

Hi everyone,
I have been trying to solve these two questions but am having trouble doing so and was wondering if someone could help me out.

(1) You wish to make a 0.200 mol/L PIPES buffer (PIPES:
1,4-Piperazine-N,N'-bis(2-ethane-sulfonic acid), C8H18N2O6S2, pKa = 6.80) at a
pH of 6.300. Available to you is the disodium salt of PIPES (Na2C8H16N2O6S2),
and a 2.00 F solution of HCl. Calculate

i)the mass of PIPES,

ii) the volume of HCl,

that must be added to make 500.0 mL of this buffer.

I tried to initally use the Henderson Hasselbach equation to solve for the ratio of weak base and its conjuate acid that would be needed to make the buffer with the desired pH but this only gets me so far. As HCL must be added i figured that I would try setting the # of moles of base in the H.Hasselbach equation to .200 + x since more moles of base will be needed to help counteract the fact that HCl is being added. However, I end up with an equation that just gives me back the original ratio of weak base and its conjugate acid, which is not correct. I have also tried writing out an acid base equation in an attempt to see what is happening with the number of moles, but I am ending up nowhere. If someone out there could please give me a hand or show me some pointers on how to solve this problem, that would be greatly appreciated.


2.
A chemist is following a hydrolysis reaction with an enzyme. The enzyme has an
optimal working pH range of 7.400.05. It is known that H+ is evolved during
the hydrolysis reaction. The chemist does a quick calculation and determines
that the reaction she is following could generate up to 0.001 moles of H+ per
100.0 mL of solution. Describe how to make up 500.0 mL of the HEPES buffer
(HEPES: 4-(2-Hydroxyethyl)piperazine-1-ethanesulfonic acid, C8H18N2O4S, pKa =
7.55) such that the reaction mixture will maintain a pH that is within the
optimal working range of the enzyme. Among the reagents in the lab are HEPES as
the free acid form, and a 5.695 F solution of NaOH. What mass of HEPES and what volume of NaOH should you use?

Note: while several solutions are possible, here you are asked to calculate the
minimum amount of HEPES that could still maintain the pH within the desired
limits.

I have tried to set up an acid base reaction to solve for the number of moles of acid are being created, but when I solve for the pH, it ends up being outside the range of the enzyme, and while I realize that one should try and minimize the amount of Hepes being used, I can't set up a mathetmatical equation that will account for this, without thinking of derivatives and setting them equal to zero to find the minimum. Could someone please give me a bit of help?


(1) You wish to make a 0.200 mol/L PIPES buffer (PIPES:
1,4-Piperazine-N,N'-bis(2-ethane-sulfonic acid), C8H18N2O6S2, pKa = 6.80) at a
pH of 6.300. Available to you is the disodium salt of PIPES (Na2C8H16N2O6S2),
and a 2.00 F solution of HCl. Calculate

i)the mass of PIPES,

ii) the volume of HCl,

that must be added to make 500.0 mL of this buffer.

I think you were headed in the right direction. You just stopped too soon.
Use the HH equation.
pH = pK2 + log (base/acid)
plug in pH of 6.3 and pk of 6.8 and solve for the base/acid ratio which you said you had done.
Then (acid) = (base)/0.316.Check my work. I may have made an arithmetic error to get the 0.316.
Then Na2P + HCl ==> NaCl + NaHP
Since we want the Na2P to be 0.2M, we use the ratio above like so,
(HP^-) = (P^-2)/0.316 = 0.2/0.316 = 0.632 M. In 500 mL this would be 0.316 mols.
Na2P + HCl ==> NaCl + NaHP
If NaHP is 0.316 mols/500 mL then
HCl must be 0.316 mols/500 mL and
HP must be 0.316 + 0.1 = 0.416 at the beginning. The HCl will use 0.316 and you will be left with 0.1 mol/500 mL which is 0.2M and that is what the problem wanted.
We can check to see if we are ok.
pH = pK2 + log(base)/(acid)
pH = 6.8 + log[(0.1/0.5)/(0.316/0.5)]
pH = 6.3.
Check my thinking and check my arithmetic. I haven't done the HCl but you know you want 0.316 mols of the 2 F solution, then add water to make the 500 mL. I hope this helps. After seeing how to do this problem the next one may be easier for you. Try it. If you still get stuck, repost that problem and tell us what the problem is.


I am sure you can calculate the grams PIPES needed to make 0.416 mols in the 500 mL so I have left the HCl and PIPES for you. I think I remember using 158 mL of the acid but check me out on that.


with water added to make the difference up to 500 mL.


I find that the mass of Pipes needed is 125.78 g. This is using a Molecular Weight for Pipes (C8H16N2O6S2) which is 302.36272 g. Would you agree that this is the molecular weight i should be using to solve for the mass of pipes, or does the molecular weight included the 2 or even 1 of the Na's? To the correct number of significant figures, I believe the answer should be given as 1.3E2 g, provided my MW is correct. I then find that volume of HCl needed is 158 mL which I also believe to have 2 sig figs and I input it as 1.6E2 mL. However, the program that I am inputting this into spits out that these two answers are incorrect. I checked over your arithmetic and it is correct, but I can't seem to think what might be wrong, unless my Molecular weight is wrong? Any ideas?


The molar mass is wrong. I think I have discarded my scratch sheets but I think I remember the molar mass of the disodium salt (that is the solid you are adding) as 346.33. Check me out on this. 158 mL HCl is what I obtained for the HCl so I think that part of your answer is correct.


I quickly read through the original problem again and I didn't see anything with less than 3 or 4 significant figures. I wonder if you can use 3 and not round to two places?


Using the HH equation.
pH = pKa + log (base/acid) . I chose pH of 7.45 because I want to minimize the amount of Hepes I have to use so making the pH higher will mean less HEPES would have had to be added originally..
plug in pH of 7.45 and pka of 7.55 and solve for the base/acid ratio
Then (acid) = (base)/0.794.
Then HEPES + NaOH ==> NaHEPES + H2O
Since we want the NaHEPES to be 0.01M, using the ratio we get .01/.794 = .01259 M. 0.01259 + .01 = moles of HEPES = .02259 mol. .02559 mol *238.29892 g/mol = 5.38 g HEPES.
Volume of NaOH = .01259 mol / 5.695 mol/L = 2.48 mL.

Can you confirm if this is what you got? This is how I tried to solve it.

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