We're doing a lab in my class and we're supposed to make a buffer at a certain pH.
I have decided to make a buffer with a pH of 5.0 using glacial acetic acid (pure CH3COOH) and sodium acetate.
I found that the ratio of acid to base is 1.41 M: 1.0 M and now I'm stuck.
My teacher only has 17.4 M glacial acetic acid and we basically have to dilute it in 75 mL of water. And I'm confused at this point. Help me please with a step by step description on how to do this?
Are you using the Henderson-Hasselbalch equation or just Ka for CH3COOH? And show me how you obtained 1.41:1.00M. I can't get that.
(pure CH3COOH) Ka = 1.8 x 10-5
sodium acetate NaCH3COO
CH3COOH --> CH3COO + H
Ka= [H+][CH3COO-]/[CH3COOH] = 1.8 x10^-5
[H+]= 4.85
pH = pKa + log [base]/[acid]
5.0 (chosen pH of buffer) = 4.85 + log [base]/[acid]
.15 = log [base]/[acid]
10^0.15 = [1.41 M]/[1 M]
OK. But the acid:base ratio is not 1.41:1. If pH you want from the problem is 5.0, then (H^+)= 1E-5 and
Ka = (H^+)(CH3COO^-)/(CH3COOH) then
(CH3COOH)/(CH3COO^-) = (H^+)/Ka
acid/base = (1E-5/1.8E-5) = 0.555. But let me show you a much easier way to do it since you typed in the HH equation.
pH = pKa + log (base/acid)
5.0 = 4.745 + log B/A
0.255 = Log B/A
B/A = 1.80 or A/B = 0.555 but I never use either of those as a number although the calculator calculates that number during the following.
Now, how many mL of the 17.4 stuff will you use? 17.4 M x # mL used = ?? millimoles. Then plug that into the HH equation and solve for base (which will be in millimoles if you used acid in millimoles).
5.00 = 4.745 + log(mmoles base)/(mmoles acid) and solve for mmoles base. I get something like 3.13 mmoles base. Convert mmoles to moles, then to grams of CH3COONa. To prepare the buffer you measure the acetic acid with a pipet (or graduated cylinder probably is good enough), add in the grams sodium acetate, stir until dissolved and you have it. Your instructor may have put some limitations on the preparation; for example, you may know how much acetic acid is to be used.
To create a buffer with a pH of 5.0 using glacial acetic acid and sodium acetate, here is a step-by-step explanation:
1. Determine the desired volume of the buffer solution. Let's assume in this case you want to make 100 mL of buffer solution.
2. Calculate the amount of sodium acetate needed. Since the ratio of acid to base is 1.41 M:1.0 M, for every 1.41 moles of glacial acetic acid, we need 1 mole of sodium acetate.
3. Convert the molar ratio to mass ratio using the molecular weight of acetic acid (60.05 g/mol) and sodium acetate (82.03 g/mol). Since the molar ratio is 1.41:1, the mass ratio will be (60.05 g/mol)/(82.03 g/mol) x 1.41 = 1.034:1.
4. Calculate the amount of glacial acetic acid needed. Multiply the mass ratio by the mass of sodium acetate you calculated in step 3 to get the mass of glacial acetic acid needed.
5. Calculate the volume of glacial acetic acid solution required. Divide the mass of glacial acetic acid by its molarity (17.4 M) to get the volume in liters.
6. Dilute the glacial acetic acid solution. You have 17.4 M glacial acetic acid, and you want to end up with 100 mL of buffer solution. Use the dilution formula:
C1V1 = C2V2
Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
In this case, C1 is 17.4 M, V1 is the volume of glacial acetic acid solution you want to determine, C2 is 1 M (because the ratio is 1.41:1 of acid to base), and V2 is 100 mL (the desired final volume).
Rearrange the equation to solve for V1:
V1 = (C2 * V2) / C1
Substitute in the values and solve for V1 to find the volume of glacial acetic acid solution you need.
7. Once you have the volume of glacial acetic acid solution needed, add that to the 75 mL of water.
8. Weigh out the calculated mass of sodium acetate obtained in step 4 and add it to the solution.
9. Mix the solution thoroughly until the sodium acetate is dissolved completely.
By following these steps, you should be able to make a buffer solution with a pH of 5.0 using glacial acetic acid and sodium acetate.