Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: suxatchem on May 09, 2005, 11:10:28 PM
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Ok here's the question i need to do.. I'm really lost!
I have 100mM (mmol.l-1) solutions of the protonated and non-protonated forms of various chemicals that are used to produce biological buffers. The chemicals are acetic acid, MOPS (4-morpholinepropane sulphonic acid) and TRIS (2-amino-2-{hydroxymethyl}-1,3-propanediol). Calculate the volume of 100mM (mmol.l-1) of the non-protonated form of TRIS (i.e. TRIS base) that I need to add to 100ml of the 100mM (mmol.l-1) protonated (i.e. TRIS.HCl) form to prepare a buffer with a pH of 7.89. Express your answer to the nearest ml. Do not include the units.
CHEMICAL FORMULAWEIGHT pKa
Acetic Acid 60 4.80
MOPS 209 7.20
TRIS 121 8.30
I believe you need to use the H-H equation but I think I'm just getting overwhelmed by the information provided, if someone could sorta outline the steps it would be really great!
Cheers
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Readyourquestion carefully - it is stated which chemicals you have to use. Just remove anything else from the question (and your thinking :) ) and use HH equation.
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ok, well this is what I've done but it got marked wrong (its one of those online quizzes, which unfortunately you get NO feedback on).
pH = pKa + log base/acid
7.89 - 8.3 log B/A
-0.41 = log B/A
10^0.41 = B/A
0.389 = base/0.1L
=0.0389L (38.9L)
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*(38.9mL)
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Express your answer to the nearest ml.
What have you entered as an answer? 38.9?
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yeah.. i just did a typo before hence the second post.
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yeah.. i just did a typo before hence the second post.
You still didn't get it ;)
38.9 is not ROUNDED to the nearest mL. Check if 39 is not a proper answer.
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I worked it out.
Should have been 10^ .41 (not -.41)
that would give, 2.57 x .1
.26L or 257mL
I had to ask lots of people, so yeah thanks heaps for not reading what i had done other than the end answer.
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Something is wrong IMHO.
39 mL gives pH = 7.89
100 mL gives pH = 8.30
257 mL gives pH = 8.71
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I worked it out.
Should have been 10^ .41 (not -.41)
that would give, 2.57 x .1
.26L or 257mL
Excuse me but it seems to me that the volume you need is 2571 mL
[H+] = 1.288E-8
Ka = 5.012E-9
1.288E-8 = 5.012E-9 * (0.1/x)
x = 0.0389 M
0.0389 mol : 1 L = 0.1 mol : y L
y = 2.57069 L => 2571 mL
That's what I guess.
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x = 0.0389 M
So far so good - assuming concentration of acid form is 0.1M, that's the concentration of base you need in the final solution.
0.0389 mol : 1 L = 0.1 mol : y L
Here I get lost - could you explain what are you trying to do here?
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In the beginning we have 0.1 L where the protonated form is 0.1 M, that's to say we would have 0.1 mol, if the volume was 1 L. But it wasn't.
Using the HH equation we find the non-protonated form is 0.0389 M. But we know we must use a solution which contains 0.1 mol of TRIS, and not 0.0389 mol.
So, we have to use a volume of TRIS 0.1M that contains the same quantity of mol in order to fit the ratio given by the HH expression. I used the following ratio:
0.0389 mol is to 1L as 0.1 mol is to 2.571L
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In the beginning we have 0.1 L where the protonated form is 0.1 M, that's to say we would have 0.1 mol, if the volume was 1 L. But it wasn't.
It was 0.01 mole of protonated form.
Using the HH equation we find the non-protonated form is 0.0389 M.
Needed nonprotonated TRIS concentration is 0.0389M. Agreed.
But we know we must use a solution which contains 0.1 mol of TRIS, and not 0.0389 mol.
If you will use a solution containing 0.1 mol of unprotonated TRIS you will end with the final solution containing 10 times more unprotonated TRIS (0.1 mol) than protonated TRIS (0.01 mol) - so you are wrong here. Do you mean 'We have to use 0.1M unprotonated TRIS solution'?
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Do you mean 'We have to use 0.1M unprotonated TRIS solution'?
Yes, you're right. What we have to do is to take the 0.1 M solution of unprotonated TRIS and water down it (or dilute it, sorry I don't know which is the best way to express it in English) up to a volume of 2.57L.
That sounds also rather similar to what suxatchem said in his/her last message.
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Yes, you're right. What we have to do is to take the 0.1 M solution of unprotonated TRIS and water down it (or dilute it, sorry I don't know which is the best way to express it in English) up to a volume of 2.57L.
OK, I think I know what the problem is. I will use HTRIS abbreviation for protonated TRIS, it will make writing shorter.
You don't have to dilute 0.1M TRIS. You have to add such an amount of this solution to the buffer that the final ratio of concentrations will be right. HTRIS concentration is not constant at 0.1M - adding 2.57L of TRIS you will dilute HTRIS already present in the solution and you will end with a buffer of some bizarre pH.
Necesary ratio of concentrations is
TRIS/HTRIS = 0.389
However - and that's were we both did a mistake - it doesn't mean that we need 0.0389M concentration. Adding TRIS you dilute HTRIS - but ratio of the concentrations will be always the same as ratio of moles is, due to the fact that volume cancels out.
So it is enough to add 0.00389 mole of TRIS - there is already 0.01 mole of HTRIS and the ratio will be OK.
That means we need exactly 38.9 mL of 0.1M TRIS solution.
I did the calculations on mole numbers from the very beginning so I didn't bother with dilution, that's probably why I have overlooked it :-[
suxatchem: could you please confirm what was the final result? I am more and more convinced that 39 mL is the right answer.
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Wheter suxatchem will answer or not, you are right.
It might have been just a typing mistake.
Due to my experience with e-exercises, I would type .039L.
Anyhow, the old exams, with the writing on a paper, are still the best.
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Cheers for the discussion guys.
I can't tell you whether that was the right answer that you came up with as the quiz just gives you a grade! but yeah im pretty sure i got that one wrong. ( i pretty sure i put in 40ml but i cant check whether that was the exact number of decimal places i entered.
The way i did it the next time, see my previous post, (with a re-worded version of the question) I did manage to get it right (as i got 90% so i couldnt have got a 2point question wrong).
But yeah hopefully what you guys have said will help me for next time. The quiz is due today, lucky you can do it as many times as you want! I only needed 80% this time but what the heck im gonna try for 100%.
Cheers. Crystal.
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this is what i was shown to do and i got it right that time;
Notice my new question is only slightly different:
Question 9: (2 points)
I have 100mM (mmol.l-1) solutions of the protonated and non-protonated forms of various chemicals that are used to produce biological buffers. The chemicals are acetic acid, MOPS (4-morpholinepropane sulphonic acid) and TRIS (2-amino-2-{hydroxymethyl}-1,3-propanediol).
Calculate the volume of 100mM (mmol.l-1) of the non-protonated form of TRIS (i.e. TRIS base) that I need to add to 100ml of the 100mM (mmol.l-1) protonated (i.e. TRIS.HCl) form to prepare a buffer with a pH of 8.59.
Express your answer to the nearest ml. Do not include the units.
CHEMICAL FORMULA
WEIGHT pKa
Acetic Acid 60 4.80
MOPS 209 7.20
TRIS 121 8.30
Answer:
pH = pKa + log [A-]/[AH]
pH - pKa = log [A-]/[AH]
8.59 - 8.3 = log [A-]/[AH]
0.29 = log [A-]/[AH]
10^0.29= [A-]/[AH]
1.95 = [A-]/[AH]
1.95 = [A-]/0.1L
1.95*.1=0.195L
=195mL
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To the OP
Ok here's the question i need to do.. I'm really lost!
I have 100mM (mmol.l-1) solutions of the protonated and non-protonated forms of various chemicals that are used to produce biological buffers. The chemicals are acetic acid, MOPS (4-morpholinepropane sulphonic acid) and TRIS (2-amino-2-{hydroxymethyl}-1,3-propanediol). Calculate the volume of 100mM (mmol.l-1) of the non-protonated form of TRIS (i.e. TRIS base) that I need to add to 100ml of the 100mM (mmol.l-1) protonated (i.e. TRIS.HCl) form to prepare a buffer with a pH of 7.89. Express your answer to the nearest ml. Do not include the units.
CHEMICAL ???FORMULAWEIGHT???pKa
Acetic Acid???60???4.80
MOPS???209???7.20
TRIS???121???8.30
I believe you need to use the H-H equation but I think I'm just getting overwhelmed by the information provided, if someone could sorta outline the steps it would be really great!
Cheers
I performed a lab similar to this during the semester, gen chem II lab.
here's the general idea, try to follow...
-use the HH equation to find the ratio [Tris base]/[Tris acid form] this is the ratio of concentrations.
-thus you are provided with the volume of the acid as well as its concentration.
100mMV1=(V1+100mL)Mb, V1 is the volume of the Tris base (non-protonated form), Mb is the final concentration of the Tris base
100mL(100mM)=(V1+100mL)Ma, Ma is the final concentration of the acid
Thus
[Tris base]/[Tris acid form] = Mb/Ma=
[100mMV1/(V1+100mL)]/[100mL(100mM)/(V1+100mL)]=[Tris base]/[Tris acid form]
The left side of the equation becomes V1/100mL, thus
V1/100mL=[Tris base]/[Tris acid ], solve for V1 and there's your answer. You might want to check up on the math also.