[Albert]

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.

[suxatchem]

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.

[suxatchem]

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

[GCT]

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.