The Kb for an amine is 2.105 × 10-5. What percentage of the amine is protonated...

The Kb for an amine is 2.105 × 10-5. What percentage of the amine 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.)

Expert Solution

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 %

queen_honey_blossom answered 2 years ago

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