Quinine has a tertiary amine group so acts as base
QN + H2O -----> QNH+ + OH-
given pka is 8.52
-log ka=8.52
ka= 3*10^-09,
for conjugate acid base pair at 25degrees Celsius kw= ka*kb
therefore Kb= kw/ka
= 10^-14/ 3*10^-09 = 3.4 *10^-06
from equation kb=[QNH+][OH-]/[QN]
QN + H2O -----> QNH+ + OH-
initial 0.018
equlli 0.018-x x x
since quinine is a weak base 0.018-x can be written as 0.018
therefore kb= x^2/0.018
x= squre root of ( 3.4*10^-6*0.018)
=2.47 *10^-04
POH= -log[OH-]= -logx= -log 2.47*10^-4
=3.61
PH= 14-POH =10.39
percent protonation= [QNH+]/Qn
=x/0.018 *100
=1.4%
Calculate the pOH, pH, and percent protonation of solute in the following aqueous solution.
0.059 M quinine, given that the pKa of its conjugate acid is 8.52.
Please show me how to do it and not just the answers. Thanks!
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1 answer