The Kb for an amine is 2.105 ×
The Kb for an amine is 2.105 × 10-5. What percentage of theamine is protonated if the pH of a solution of the amine is 9.788?(Assume that all OH– came from the reaction of B with H2O.)
Answer:
pH = 9.788
pH + pOH = 14
pOH = 14 – 9.788
pOH = 4.212
[OH-] = 10^-4.212 = 6.14 x 10^-5 M = x
Kb = 2.105 × 10^-5
B + H2O ———————> BH+ + OH-
C-x x x
Kb = x^2 / C – x
2.105 × 10^-5 = (6.14 x 10^-5 )^2 / C – 6.14 x 10^-5
C – 6.14 x 10^-5 = 1.77 x 10^-4
C = 2.38 x 10^-4 M
amine protonated = ( x / C ) x 100
= (6.14 x 10^-5 / 2.38 x 10^-4 ) x 100
= 25.78 %
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