At 22 °C, an excess amount of a generic metal hydroxide, M(OH)2, is mixed with pure water. The resulting equilibrium solution ha
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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⁻
<em>Where A is neutral base and AH⁺ is protonated form</em>
The basic dissociation constant, kb, is:
As pH in stomach is 2,5:
[OH] =
[OH] = 3,16x10⁻¹² M
Thus:
<em>(1)</em>
Using (1) it is possible to know if you have the neutral base or the protonated form, thus:
(a) nicotine Kb = 7x10^-7
[AH⁺}>>>>[A]
For nicotine, the protonated form is the present in stomach
(b) caffeine,Kb= 4x10^-14
[AH⁺}<<<<[A]
For caffeine, the neutral base form is the present in stomach
(c) strychnine Kb= 1x10^-6
[AH⁺}>>>>[A]
For Strychinene, the protonated form is the present in stomach
(d) quinine, Kb= 1.1x10^-6
[AH⁺}>>>>[A]
For quinine, the protonated form is the present in stomach.
I hope it helps!