I suppose the problem is asking for the anion. If HB is the acid, then the base is B^- and that hydrolyzes in water solution this way.
.............B^- + HOH ==> HB + OH^-
I..........0.1M............0....0
C...........-x.............x....x
E.........0.1-x............x....x
K for the base is not listed but it can be calculated as Kbase = Kw/Kacid = 1E-14/2.82E-8
Then Kbase = (HB)(OH^-)/(HB)
2.81E-8 = (x)(x)/(0.1-x)
Solve for x = OH^- and convert to pH.
If you make up a solution of 100mL of 0.1 M Hepes in the basic form, what will be the pH? pKa = 7.55
3 answers
Dr. Bob...I have seen you trying to post answers to several questions, but in reality all you are doing is confusing everyone more. Please be more straight forward with your answers. In your other posts, I have seen that several people can not follow what you are saying.
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