[d.beser]

Hi all!

I am to do a lab that requires me to figure out what buffer system I am to use with an assigned pH. In my case I was assigned with a pH=7.4.

The buffer systems are as follows:
A) 3.5M acetic acid                    pKa = 4.74                     molecular weight=60.1
    sodium acetate, anhydrous                                          molecular weight= 82.0

B) sodium phosphate, monobasic            pKa=6.86          molecular weight=138.0
    sodium phosphate, dibasic                                           molecular weight=142.0


I did the caclulations to figure out ratio...

pH=pKa+log[(b)/(a)]
A) 7.4=4.74+log [(b)/(a)]                                     B) 7.4=6.86+log [(b)/(a)]
      457.0881896=[(b)/(a)]                                        3.5=[(b)/(a)] 
      457.1:1                                                               3.5:1

After figuring out ratios for both Buffer systems I then calculated their percentages of base and acid form

A) (457.1)/(457.1+1)= 0.998 or 99.8% in base form
    (1)/(457.1+1)= 0.0023 or 0.23% in acid form

B) (3.5)/(3.5+1)= 0.78 or 78% in base form
    (1)/(3.5+1)= 0.22 or 22% in acid form

Based on this information and my assigned pH=7.4 I know I am to use buffer system B because of the pKa value. I am to figure out how much(amonunt) weak acid and conjugate base I need to make when I use 25mL of 0.25M acid solution and 25mL of 0.25M of base solution.

Up until the percent obtained everything is correct, but after that the figuring out the amount I did incorrectly!

What I did incorrectly is:

1)25mL=0.025L                                                           
   0.25M

0.25Mx0.025L= 0.00625 mol

0.00625x0.78 = 0.0049 mol
0.00625x0.22 = 0.0014 mol

0.0049x138.0(mw)= 0.68 grams
0.0014x142.0(mw)= 0.20 grams


2) 50mL=0.050L
    0.50M

0.50Mx0.050L=0.025mol

0.025x0.78 = 0.0195mol
0.025x0.22= 0.0055mol

0.0195x138.0 = 2.691 grams
0.0055x142.0 = 0.781 grams

How do I create a solution based on the amount of base and acid in terms of just using H-H equation.
I am extremely confused!

[Borek]

In general you seem to be on the right track (could be there is some subtle mistake, I have just skimmed), but I don't understand why you use different concentrations for 25 mL and 50 mL solutions.

How much of the buffer solution do you have to prepare, and what materials do you have to use?

[d.beser]

Thank you on your reply!

Thats where I am stuck.

The question is to choose the buffer system which is most appropriate to the assigned pH. I chose buffer system B because of its pKa value.

Then I am to caclulate the amounts of weak acid and conjugate base (one acid solution is 25mL of 0.25M and one base solution is 25mL of 0.25M) so that I can prepare 50mL of 50mM buffer at the assigned pH using the appropriate volumes of the 0.25 solutions of buffer.

So, how do I go from finding ratio to finding how much acid and base I will be using?

[Babcock_Hall]

Aa a practical matter, I would be tempted to break the process of making up such a buffer into two steps.  In the first I would prepare a concentrated buffer with the correct ratios of acid to base.  In the second I would dilute an appropriate volume of this stock solution to the final desired volume.  One could criticize such a procedure on a couple of grounds, however.

[Borek]

Then I am to caclulate the amounts of weak acid and conjugate base (one acid solution is 25mL of 0.25M and one base solution is 25mL of 0.25M) so that I can prepare 50mL of 50mM buffer at the assigned pH using the appropriate volumes of the 0.25 solutions of buffer.

Are you sure that's what you are expected to do? You can't prepare a solution of a concentration higher than the concentrations of solutions used.

Besides, the way you worded your post seems like you are given 25 mL of both solutions and you are expected to prepare 50 mL of the target solution - so you have to use everything (and it definitely doesn't leave you with a much choice of the final solution pH).

[Babcock_Hall]

Borek,

I may be wrong, but I think that the final buffer is supposed to be 50 mM total buffer but the initial solutions are supposed to be made up at 250 mM.

[d.beser]

The initial volume is 0.025L and the initial concentration is 0.25M

how do I calculate grams of acid and base needed to make the buffer?

Acid=7.76grams
Base=2.24grams

How does one get this answer?

[Babcock_Hall]

When you say "initial," what exactly do you mean?  How is it distinguished from "final?"

[d.beser]

Both acid and base have a initial concentration of 0.25M and a initial volume of 25mL. Their final concentration is 0.50M and a final volume of 50mL.

Base has a percentage of 78%                                              ratio is 3:1
Acid has a percentage of 22%

I am to find in grams how much base and acid is need to make the buffer so that its establishes a pH of 7.4

[d.beser]

Choose the buffer system most effective at pH 7.4

Buffer System:                  Chemicals:                                     pKa:                             M.M
-------------------------------------------------------------------------------------------------------------------
A                                      3.5M acetic acid                             4.74                              60.01
                                        Sodium acetate, anhydrous                                                 82.0

B                                      sodium phosphate, monobasic          6.86                            138.0
                                        sodium phosphate, dibasic                                                  142.0
--------------------------------------------------------------------------------------------------------------------

pH=pKa+log[(b)/(a)]

7.4=6.86+log [(b)/(a)]
3.5=[(b)/(a)]
3.5:1

(3.5)/(3.5+1)= 0.78 or 78% in base form
(1)/(3.5+1)= 0.22 or 22% in acid form

-----------------------------------------------------------------------------------------------------------------------------------------------------
Calculate the amounts of weak acid and conjugate base needed to make 25mL of 0.25M solutions (one acid solution and one base solution).

..I dont know how to calculate amounts of weak acid and conjugate base needed! Any help would be greatly appreciated!


                                     

[Arkcon]

d.beser:, I hope you don't mind my merging this new post with your previous post on a similar topic.  It will be useful for us to see what you knew only a few days ago.  Can you begin to start this new problem with what's been talked about already?

[d.beser]

I don't mind at all! I'm just not finding the help I need here. Is there anyone who would know how the answers acid=7.76g and base=2.24g were obtained. I've been stuck on this question for a week and have been looking everywhere for examples.

[d.beser]

Choose the buffer system most effective at pH 7.4

Buffer System:                  Chemicals:                                     pKa:                             M.M
-------------------------------------------------------------------------------------------------------------------
A                                      3.5M acetic acid                             4.74                              60.01
                                        Sodium acetate, anhydrous                                                 82.0

B                                      sodium phosphate, monobasic          6.86                            138.0
                                        sodium phosphate, dibasic                                                  142.0
--------------------------------------------------------------------------------------------------------------------

pH=pKa+log[(b)/(a)]

7.4=6.86+log [(b)/(a)]
3.5=[(b)/(a)]
3.5:1

(3.5)/(3.5+1)= 0.78 or 78% in base form
(1)/(3.5+1)= 0.22 or 22% in acid form

-----------------------------------------------------------------------------------------------------------------------------------------------------
Calculate the amounts of weak acid and conjugate base needed to make 25mL of 0.25M solutions (one acid solution and one base solution).

..I dont know how to calculate amounts of weak acid and conjugate base needed! Any help would be greatly appreciated!

[mjc123]

I've no idea how those numbers were obtained. They look suspiciously similar to your 78% and 22% - but that was for molarity, not mass.
You have correctly calculated that 25 ml of 0.25M is 0.00625 mol. But that would be less than a gram of acid or base.
Besides, there is confusion in your statements. You say

Calculate the amounts of weak acid and conjugate base needed to make 25mL of 0.25M solutions (one acid solution and one base solution).

but you also say

I am to find in grams how much base and acid is need to make the buffer so that its establishes a pH of 7.4

Which is it - the stock solutions or the final buffer?
Can you quote (or post) the original question exactly as it was asked?

[Arkcon]

OK.  So what's the remaining question?  You've used Henderson-Hasselbach correctly.  You've determined relative amounts of acid and base to use.  You don't know what to do next?

Consider, you have to make a buffer.  How much buffer do you have to make?  One liter?  Ten liters?  Or 100 mL?  How strong?  Is it 1M, or 0.1M or 25 mM?  You've got the proportions, but your work isn't done, not if you want to actually make a solution to use in an experiment.