Answer:
a) For nicotine, the protonated form is the present in stomach.
b) For caffeine, the neutral base form is the present in stomach.
c) For Strychinene, the protonated form is the present in stomach.
d) For quinine, the protonated form is the present in stomach.
Explanation:
In a basic dissociation for molecules with basic nitrogen, the equilibrium is:
A + H₂O ⇄ AH⁺ + OH⁻
Where A is neutral base and AH⁺ is protonated form
The basic dissociation constant, kb, is:
[tex]K_{b} = \frac{[AH^+][OH^-]}{[A]}[/tex]
As pH in stomach is 2,5:
[OH] =[tex]10^{-[14-pH]}[/tex]
[OH] = 3,16x10⁻¹² M
Thus:
[tex]\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}[/tex] (1)
Using (1) it is possible to know if you have the neutral base or the protonated form, thus:
(a) nicotine Kb = 7x10^-7
[tex]\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}[/tex]
[tex]\frac{7x10^{-7}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}[/tex]
[tex]221359 = \frac{[AH^+]}{[A]}[/tex]
[AH⁺}>>>>[A]
For nicotine, the protonated form is the present in stomach
(b) caffeine,Kb= 4x10^-14
[tex]\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}[/tex]
[tex]\frac{4x10^{-14}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}[/tex]
[tex]0,0127 = \frac{[AH^+]}{[A]}[/tex]
[AH⁺}<<<<[A]
For caffeine, the neutral base form is the present in stomach
(c) strychnine Kb= 1x10^-6
[tex]\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}[/tex]
[tex]\frac{1x10^{-6}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}[/tex]
[tex]314456 = \frac{[AH^+]}{[A]}[/tex]
[AH⁺}>>>>[A]
For Strychinene, the protonated form is the present in stomach
(d) quinine, Kb= 1.1x10^-6
[tex]\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}[/tex]
[tex]\frac{1,1x10^{-6}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}[/tex]
[tex]347851= \frac{[AH^+]}{[A]}[/tex]
[AH⁺}>>>>[A]
For quinine, the protonated form is the present in stomach.
I hope it helps!
The ratio of [AH^+]/[A] determines the state of the substance in the stomach at that pH.
Consider the generic reaction;
A + H2O ⇄ AH^+ + OH^-
Let A be the neutral base and AH^+ is its protonated form
Let us recall that;
Kb = [AH^+] [OH^-]/[A]
Hence;
Kb/[OH^-] = [AH^+]/[A]
Note that the pH of the stomach is 2.5
H^+ = Antilog(-2.5) = 3.16 × 10^-3
OH^- = 1 × 10^-14/3.16 × 10^-3
OH^- = 3.16 × 10^-12
For nicotine;
7x10^-7/3.16 × 10^-12 = [AH^+]/[A]
[AH^+]/[A] = 2.2 × 10^5
Since [AH^+]>>>[A], nicotine will exist in the protonated form
For caffeine;
4x10^-14/3.16 × 10^-12 = [AH^+]/[A]
[AH^+]/[A] = 1.27 × 10^-2
Since [AH^+]<<<<<[A], caffeine will exist in the neutral form
For strychnine
1x10^-6/3.16 × 10^-12 = [AH^+]/[A]
[AH^+]/[A] = 3.16 × 10^5
Since [AH^+]>>>[A], strychnine will exist in the protonated form
For quinine
1.1x10^-6/3.16 × 10^-12 = [AH^+]/[A]
[AH^+]/[A] = 3.48 × 10^5
Since [AH^+]>>>[A], quinine will exist in the protonated form
Learn more about protonated form:https://brainly.com/question/731147
