[suxatchem]

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

[Borek]

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.

[suxatchem]

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)          

[suxatchem]

*(38.9mL)

[Borek]

Express your answer to the nearest ml.

What have you entered as an answer? 38.9?

[suxatchem]

yeah.. i just did a typo before hence the second post.

[Borek]

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.

[suxatchem]

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.

[Borek]

Something is wrong IMHO.

39 mL gives pH = 7.89
100 mL gives pH = 8.30
257 mL gives pH = 8.71

[Albert]

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.

[Borek]

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?

[Albert]

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

[Borek]

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'?

[Albert]

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.

[Borek]

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.