I can get you started.
For pH = 5.00 start with the HH equation and solve for base/acid ratio.
5.00 = 4.74 + log(b/a)
b/a = about 1.82 or
base = 1.82*acid (equation 1)
base + acid = 20 millimoles.
Solve those two equations simultaneously. I get something like 13 for base and 7 for acid but those are approximate. You need to go through and obtain better numbers.
You want 100 mL 0.2M buffer which is 20 millimols. I would start with 0.2 x 82 = about 16.4 g sodium acetate and this i to be divided up as 20
.......Ac^- + H^+ ==> HAc
You want x to be 7; therefore, 20-x = 13. Convert those to mL 11.7M HCl to be added and I'll leave the dilutions to you. You should go through a quick check to make sure what you're preparing actually is pH = 5.00
pH = 4.74 x log(13/7) = 5.00
Dr. Bob222, Iím posting this because I was working on it on a word document and had to go and take care of something before I was able to finish the dilution part; I didn't want to do this much work and not be able to post it. I am at least glad that we were on the same page on how to tackle this problem.
NaC2H3O2 + HCl --> HC2H3O2 + NaCl
You will need the following equation:
Solving for [A-]/[HA]
Which should give you a ratio of 1.74
You need 0.2 M of the buffer, so you need (100mL)*0.2M = 0.02 moles of sodium acetate/acetic acid
0.02= A- + HA
Substituting one equation into the other, 1.74=0.02-HA/HA
Solving for HA, which is the amount of HCl needed, gives 0.00730 moles of HCl; this is so because HCl completely dissociates.
0.02 moles of sodium acetate*(82 g/1 mole)= mass of sodium acetate.
11.7N HCl=11.7M HCl
(0.00730 moles of HCl/11.7M)*10^3= volume of HCl in mL
100mL-volume of HCl in mL= additional volume of solvent needed.
For the dilutions just take part of the solution and dilute it with solvent i.e., 1 part solution and 4 parts solvent for the 1:4 dilution; repeat as indicated from the stock solution that you made.
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