Posted by bob on Tuesday, April 24, 2012 at 9:18pm.
Let's call quinine, BN
.........BN + HOH ==> BNH^+ + OH^-
initial.0.39M.........0........0
change...-x...........x.........x
equil...0.39-x........x........x
Kb = (BNH^+)(OH^-)/(BN)
You have pKa of the conjugate acid, convert to pKb for the base, substitute into the Kb expression and solve for (OH^-) then convert that to pH.
%protonation = [(OH^-)/(BN)]*100 = ?
-log ka=8.52
ka= 3*10^-09
Kb= kw/ka
= 10^-14/ 3*10^-09 = 3.4 *10^-06
kb= x^2/0.39
x= squre root of ( 3.4*10^-6*0.39)
=1.15*10^-3
POH= -log[OH-]= -logx= -log 1.15*10^-3
=2.939
PH= 14 - 2.939
=11.06
IS THIS RIGHT?
nvm, it IS right, ty xD
I calculate 3.31E-6 for Kb which runs through the rest of the problem and ends with 11.055 which I would round to 11.06.
for the %protonation = [(OH^-)/(BN)]*100
i tried doing 1.15*10^-3 / 0.39 * 100
= .295, but the answer is incorrect.
can you tell me what i did wrong?
I think it goes back to the incorrect conversion of pKa to pKb to Kb. Then for OH^- I obtained 1.136E-3.
0.001136/0.39 = 0.00291 and that times 100 = 0.291% which rounds to 0.29 to 2 significant figures.
Your problem may very well be that you are reporting too many s.f. I'll bet 0.29% will work. These data bases are "good" about that and catch a lot of trouble for it. Let me know.
YES! it was the OH^- that messed up my calculation. The answer you provided me was correct, thanks for all the help!
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