Posted by dIrk on Friday, January 21, 2011 at 7:32pm.
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?
biochemistry - DrBob222, Friday, January 21, 2011 at 7:51pm
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.
biochemistry - dIrk, Friday, January 21, 2011 at 7:58pm
what's the pKa?
biochemistry - dIrk, Friday, January 21, 2011 at 8:04pm
pH= pKa + log 0.1M/0.1M
what's the pKa value?
biochemistry - DrBob222, Friday, January 21, 2011 at 8:07pm
Won't that be pKa for k2 of H3PO4?
The 9.44 is pKa for ethanolamine.
biochemistry - dIrk, Friday, January 21, 2011 at 8:09pm
biochemistry - DrBob222, Friday, January 21, 2011 at 8:29pm
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).
biochemistry - dIrk, Sunday, January 23, 2011 at 6:51pm
pH=7.20+log 0.2M/ 0.4M
I'm heading in the right direction?
biochemistry - dIrk, Monday, January 24, 2011 at 5:09pm
Sorry the final pH of the mixture is 7.50, is that correct DrBob222?
biochemistry - dIrk, Monday, January 24, 2011 at 5:12pm
the equation will look like this:
biochemistry - Jimmy, Saturday, February 5, 2011 at 12:29am
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!!!
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