Posted by **Abdullateef ** on Saturday, March 16, 2013 at 8:28am.

perform the calculation required for100ml of 0.2 M sodium acetate buffer of pH 5.0. If you are given a bottle of sodium acetate acetate(mw =82 and a bottle of concentrated HCl of normality =11.7). After the calculation, prepare the buffer , check the pH of your calculation and comment on your result . Prepare 1:1, 1:4, 1:10, and 1:100 diluting your buffer and obtained pH reading for them

- biochemistry -
**DrBob222**, Saturday, March 16, 2013 at 2:35pm
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

I......20.......0.......0

add............x.........

C.....-x......-x.........x

E.....20-x....x.........7

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

- biochemistry -
**Devron**, Saturday, March 16, 2013 at 5:46pm
Dr. Bob222, I’m 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:

pH=pka+log([A-]/[HA])

Where

pH=5

pKa=4.76

Solving for [A-]/[HA]

10^(pH-pka)=[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

And

1.74=[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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