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}]}](https://tex.z-dn.net/?f=kb%20%3D%20%5Cfrac%7B%5BRNH_%7B4%7D%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BNH_%7B3%7D%5D%7D)
As kb = 4,004x10⁻⁵ and [OH⁻] is
:
![4,004x10^{-5} = \frac{[RNH_{4}^+][2,34x10^{-5}]}{[NH_{3}]}](https://tex.z-dn.net/?f=4%2C004x10%5E%7B-5%7D%20%3D%20%5Cfrac%7B%5BRNH_%7B4%7D%5E%2B%5D%5B2%2C34x10%5E%7B-5%7D%5D%7D%7B%5BNH_%7B3%7D%5D%7D)
1,708 = [RNH₄⁺] / [RNH₃] <em>(1)</em>
As the total amine is a 100%:
[RNH₄⁺] + [RNH₃] = 100% <em>(2)</em>
Replacing (1) in (2):
1,708 [RNH₃]+ [RNH₃] = 100%
2,708 [RNH₃] = 100%
[RNH₃] = 36,93%
Thus,
<em> [RNH₄⁺] = 63,07%</em>
The percentage of amine protonated is 63,07%
I hope it helps!