biochemistry
posted by dIrk on .
if 4 volumes of 0.1M monobasic potassium phosphate, KH2PO4, are mixed with 2 volume of 0.1 M dibasic sodium phosphate, Na2HPO4, what will be the pH of the mixture?
What if 4 volumes of 0.1 M KH2PO4 are mixed with 8 volumes of 0.1 M ethanolamine base(pka=9.44) what will be pH of mixture?

Use pH = pKa + log[(base)/(acid)]
KH2PO4 is the acid. Na2HPO4 is the base.
For b part:
Write the reaction between KH2PO4 (the acid) and ethanolamine(the base) and see how much of each is left over and the amount of salt produced. Then plug into the HH equation.
Post your work if you get stuck. 
what's the pKa?

pH= pKa + log 0.1M/0.1M
what's the pKa value? 
Won't that be pKa for k2 of H3PO4?
The 9.44 is pKa for ethanolamine. 
pH=7.20+log0.1M/0.1M
pH=7.20 
base is not 0.1 and acid is not 0.1. Isn't the base 4 volumes x 0.1 = 0.4 moles; therefore, the concn is 0.4 moles/volume (whatever that is but it doesn't matter because the acid will be divided by the same V and V will cancel).

pH=7.20+log 0.2M/ 0.4M
pH=7.20+(0.30)
pH=6.69
I'm heading in the right direction? 
Sorry the final pH of the mixture is 7.50, is that correct DrBob222?

the equation will look like this:
pH=7.20+log 0.4/0.2
= 7.50
Yes? 
I believe you, correct if I'm not mistaken on part A. The answer is 7.50.
for part B: the Base is the amount of initial base minus the amount of base protonated.
Maybe Dr.Bob should show us Visually . DR.BOB!!!