Question

The K b Kb for an amine is 4.004 × 10 − 5 . 4.004×10−5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.370 9.370 ? Assume that all OH − OH− came from the reaction of B with H 2 O . H2O. percentage protonated: %

Answer 1

Answer:

The percentage of amine is protonated is 63,07%

Explanation:

The reaction of an amine RNH₃ (weak base) with water is:

RNH₃ + H₂O ⇄ RNH₄⁺ + OH⁻

The kb is defined as:

$kb = \frac{[RNH_{4}^+][OH^-]}{[NH_{3}]}$

As kb = 4,004x10⁻⁵ and [OH⁻] is $10^{-(14-9,370)}=2,344x10^{-5}$:

$4,004x10^{-5} = \frac{[RNH_{4}^+][2,34x10^{-5}]}{[NH_{3}]}$

1,708 = [RNH₄⁺] / [RNH₃] (1)

As the total amine is a 100%:

[RNH₄⁺] + [RNH₃] = 100% (2)

Replacing (1) in (2):

1,708 [RNH₃]+ [RNH₃] = 100%

2,708 [RNH₃] = 100%

[RNH₃] = 36,93%

Thus,

[RNH₄⁺] = 63,07%

The percentage of amine protonated is 63,07%

I hope it helps!