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Question: Nicotine, \[{C_{10}}{H_{14}}{N_2}\], has two basic nitrogen atoms and both can react with water to g...

Nicotine, C10H14N2{C_{10}}{H_{14}}{N_2}, has two basic nitrogen atoms and both can react with water to give a basic solution
Nic(aq)+H2O(l)⇌NicH+(aq)+OH−(aq)Ni{c_{(aq)}} + {H_2}{O_{(l)}} \rightleftharpoons Nic{H^ + }_{(aq)} + O{H^ - }_{(aq)}
NicH+(aq)+H2O(l)⇌NicH2(aq)2++OH−(aq)Nic{H^ + }_{(aq)} + {H_2}{O_{(l)}} \rightleftharpoons Nic{H_2}{^{2 + }_{(aq)}} + O{H^ - }_{(aq)}

Kb1  is  7×10−7{K_{b1}}\;is\;7 \times {10^{ - 7}}and Kb2  is  10−10.{K_{b2}}\;is\;{10^{ - 10}}.Calculate the approximate pHpHof 0.020 M0.020{\text{ }}Msolution.

Explanation

Solution

Nicotine is a stimulant drug as it speeds up our reactions and processes which occur in our body. It speeds up the timing of each reaction occurring. It is a psychoactive ingredient which is present in tobacco as its main component. When tobacco is burnt, tar and carbon monoxide is also released which means it's released when tobacco is smoked.

Complete step-by-step solution: Various other products such as cigars, chewing tobacco and wet and dry snuff as well as the dried leaves consist of nicotine. Electronic cigarettes(also known as E cigarettes) do not contain dried tobacco leaves, but they may still contain nicotine.
Nicotine is a plant alkaloid and is found in the plant of tobacco. It is an addictive central nervous system stimulant which either leads to ganglionic stimulation in low doses or causes ganglionic blockages which are consumed in high doses.
Nicotine is extensively metabolized by first pass metabolism in the liver. The lung and kidneys can also metabolize nicotine but to a much smaller extent.
Thus we need to calculate the approximate pHpHof 0.020 M0.020{\text{ }}Msolution by the following steps:
Kb1=8×10−7=[NicH+][OH−]/[Nic]{K_{b1}} = 8 \times {10^{ - 7}} = [NicH + ][OH - ]/\left[ {Nic} \right]
[Nic].....[NicH+].....  [OH−]\left[ {Nic} \right].....[NicH + ].....\;[OH - ]
I          0.2                   0                             0I\;{\text{ }}\;{\text{ }}\;\;0.2\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;\;0\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;0
             −x              +x                       +x\;{\text{ }}\;{\text{ }}\;{\text{ }}\;\; - x\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\; + x\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\; + x
0.2−x             x                            x0.2 - x\;{\text{ }}\;{\text{ }}\;{\text{ }}\;\;x\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;\;x
∴8×10−7=x2/0.2−x\therefore 8 \times {10^{ - 7}} = {x^2}/0.2 - x ⇒x2+8×10−7x−1.6×10−7=0 \Rightarrow {x^2} + 8 \times {10^{ - 7}}x - 1.6 \times {10^{ - 7}} = 0
∴x=[NicH+][OH−]=0.00039=3.9×10−4M\therefore x = [NicH + ][OH - ] = 0.00039 = 3.9 \times {10^{ - 4}}M
Then to further find out,
kb2=1×10−10=[NicH22+][OH−]/[NicH+]{k_{b2}} = 1 \times {10^{ - 10}} = [Nic{H_2}^{2 + }][O{H^ - }]/[Nic{H^ + }]
[NicH+].....[NicH22+].....[OH−][NicH + ].....[NicH22 + ].....[OH - ]
0.00039                 0                    0.000390.00039\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;0\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;0.00039
      −x                      +x                       +x\;\;\; - x\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;\; + x\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\; + x
0.00039−x          x                0.00039+x0.00039 - x\;{\text{ }}\;{\text{ }}\;\;x\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;\;0.00039 + x
∴10−10=x21/0.00039     [x  in denominator can be ignored]\therefore {10^{ - 10}} = {x^2}1/0.00039\;{\text{ }}\;[x\;in{\text{ }}denominator{\text{ }}can{\text{ }}be{\text{ }}ignored]
⇒x2=3.9×10−4×10−10\Rightarrow {x^2} = 3.9 \times {10^{ - 4}} \times {10^{ - 10}}
⇒x=1.97×10−7\Rightarrow x = 1.97 \times {10^{ - 7}}
∴[OH−]=0.00039+1.97×10−7=0.00039M\therefore [OH - ] = 0.00039 + 1.97 \times {10^{ - 7}} = 0.00039M
∴pH=14−pOH\therefore pH = 14 - pOH
=14−3.4=10.6= 14 - 3.4 = 10.6
∴  Approximate  pH=10.6\therefore \;Approximate\;pH = 10.6

Note: As a result, peripheral vasoconstriction, tachycardia, and elevated blood pressure may be observed with nicotine intake. This agent may also stimulate the chemoreceptor trigger zone, thereby inducing nausea and vomiting. Nicotine is a natural alkaloid that is a major component of cigarettes and is used therapeutically to help with smoking cessation.