Chemistry Homework Solutions
#25375
Buffers
An unknown compound X, is thought to have a carboxyl group with a pka of 2.0 and another ionizable group with a pka between 5 and 8. When 75 ml of 0.1 M NaOH was added to 100 ml of a 0.1 m solution of X at pH 2.0, the pH increased to 6.72.Calculate the pka of the second ionizable group of X.
I tried to solve this using the Henderson Hasselbalch equation.
6.72=pka2+log[A-]/[HA]
6.72=pka2+log[0.10/0.075]
6.72=pka2+log(1.33)
6.72=pka2+0.12
pka2= 6.60
But this does not seem to be the right answer. I think I don't have the proper proportions
What is this?
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