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Henderson Hasselbach Eq'n. Tris HCL Buffer

  • Thread starter uzi
  • Start date
  • #1
uzi
1
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i need to create a solution of 0.70M Tris HCL buffer of pH 7.9 using 12.1 M of HCL
pKa = 8.3
so far this is what i have:

I need 8.49 g of Tris HCL because:
...(o.7M)*0.1L *(121.4g/mol)
THIS part of the calculation is right so dont worry about it. however i cant get the ratios right.

pH = pKa + log (base/acid)
7.9=8.3 + log (Trisb/Trisa)
7.9-8.3 = log (Tb/Ta)
-04 = log (Tb/Ta)
antilog -0.4 = 0.4

There are 4parts acid to 1 part base therefore:
Tb = 0.4Ta
Total tris = 0.70M
ThereforeL 0.7M =Tb +Ta
0.7 = 0.4Ta + Ta
0.7 = 0.4(4 parts Ta) + 0.4
0.7 = 2Ta

hence:

Ta = 0.7M/2 = 035
Tb = 0.70M - 0.35 = 0.35

BUT this doesnt make sense, cuz if youplug the ratios into the HH equation you dont get the right pH:

pH = 8.3 + log (0.35/0.35)
pH = 8.3 + 0
pH = 8.3
I need to make it a pH of 7.9 !!!!!!!!

and then on top of that since the ratios are wrong I cant successfully calculate the HCL volume. T.T

The only correct ratios that would make sense are .2 and .5, but I did that through trial and error therefore I cannot use it.

pH = 8.3 + log (0.2/0.5)
pH = 8.3 + (-0.397)
pH = 7.9
 

Answers and Replies

  • #2
Borek
Mentor
28,412
2,815
0.7 = 0.4Ta + Ta
0.7 = 0.4(4 parts Ta) + 0.4
0.7 = 2Ta
Check your math. First equation is OK, but later you do some unrelated tricks.

The only correct ratios that would make sense are .2 and .5
Hardly surprising: 0.2/0.5=0.4, so it just confirms what you calculated at the very beginning.
 

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