How to calculate amounts of acid and conjugate base?

Hello,

I need help with my biochem lab. Here is the question...

"Prior to lab, use the Hendersen- Hasselbalch equation to calculate the amount of
Acetic acid and acetate needed to make a 100 mL solution of 50 mM
acetate/acetic acid with a pH of 4.0. For this part 0.2 M acetic acid will be the stock
solution. Measure out the correct amounts of each compound, add water to bring the final
volume to 100 mL, and determine the pH of
the solution. Make note of your calculations
at this step and explain how you determined the amount of each reagent to get at pH of
4.0. [MW of acetic acid = 60.05 g/mol, MW of sodium acetate = 82.04 g/mol, density of
acetic acid = 1.049 g/mL, pKa of acetic acid = 4.757]  "

.....So ph = pka + log ([A-] / [HA])
          4.0 - 4.757 = log ([A-] / [HA])
          -0.757 = log ([A-] / [HA]) 
         10^(-0.757) = 10 ^ (log [A-] / [HA])
         .175 = [A-] / [HA] ----> this means that if we are dealing with a solution in a Liter volume, then there would be .175 moles of A- per mole of HA? Is my logic right?

But since we need to make a solution of 100 mL of acetate/acetic acid. There will be 0.0175 moles of A- and 0.1 moles of HA.

0.0175 moles of A- (acetate) x 82.04 g/mol of acetate = 1.436 grams of Acetate
0.1 moles of HA (acetic Acid) x 60.05 g/mol of acetic acid = 6.005 grams of Acetic Acid

6.005 grams of acetic acid x Density (1mL/1.049g) = 5.724 mL of Acetic Acid


Im not sure if I am really wrong here or if I'm close. Please help!

Post Merge: 9 years ago

I got a bit antsy and I searched the forums and may have found a better alternative....but I'm really not sure.

pH = pka + log A/HA
4 -4.757 = log A / 0.05M  - I am not sure if I am supposed to use 0.05 M or 0.2 M for HA
-0.757 = log A / 0.05M
10^ (-0.757) = 10^(log A / 0.05M)
0.175 = A / 0.5M
A- (acetate) = 0.0875 moles/ L x 0.1 L = 0.00875 moles x 82.04 g/mol = 0.712 grams of Acetate

- I'm lost from here.....

Post Merge: 9 years ago

ok, last update....I need some sleep lol...

pH = pka + log A/HA
4 -4.757 = log A / 0.2M  Left Arrow I went ahead and used 0.2 M (it seemed like the better choice)
-0.757 = log A / 0.2M
10^ (-0.757) = 10^(log A / 0.2M)
0.175 = A / 0.2M
A- (acetate) = 0.0350 moles/ L x 0.1 L = 0.0035 moles x 82.04 g/mol = 0.287 grams of Acetate

0.2 M of HA x 0.1 L = 0.02 moles x 60.05 g/mol = 1.201 grams
1.201 grams x (1 mL / 1.049 g) = 1.145 mL of Acetic Acid

Am I right? Wrong? Please let me know!

Post Merge: 9 years ago

Well, I got the problem wrong but at least I know the answer now. I'll post my work here for anybody else in the future that runs across this annyoing problem...

The molarity of Acetic acid, in buffer, is not 0.2 M or 50mM. 50 mM represent the combined molarity of Acetic acid + Acetate salt. So a relationship needs to be established between the acid and its conjugate base

[A-] + [HA] = 50 mM 0.05 M
[A-] = 0.05 M - [HA]  you must plug this value into the Hendersen-Hasselbalch equation

ph = pka + log ([A-] / [HA])
4.0 - 4.757 = log ((0.05 M - [HA]) / [HA])
1.175[HA] = 0.05 - [HA]
[HA] = 0.0425 M to convert to moles you need to multiply by how much you are using which in this case is 100 ml or 0.1 L

0.0425 moles/L (molarity) x 0.1 L = 0.00425 moles of Acetic Acid x 60.05 g/mol x 1mL/1.049 (density) = 2.433 mL of Acetic Acid

Now you can take the derived molarity of Acetic Acid to get molarity of Acetate using the established relationship equation from before.

[A-] + [HA] = 0.05 M
[A-] = 0.05M - 0.0425 M
       = 0.0075 M x 0.1L = 0.00075 moles of Acetate salt
0.00075 moles x 82.04 g/mol of sodium acetate = 0.0615 g