91Ó°ÊÓ

Nicotine undergoes two protonation reactions:

\(\begin{aligned}{{\bf{C}}_{{\bf{10}}}}{{\bf{H}}_{{\bf{14}}}}{{\bf{N}}_{\bf{2}}}{\bf{(aq) + }}{{\bf{H}}_{\bf{2}}}{\bf{O(l)}} \to {{\bf{C}}_{{\bf{10}}}}{{\bf{H}}_{{\bf{14}}}}{{\bf{N}}_{\bf{2}}}{{\bf{H}}^{\bf{ + }}}{\bf{(aq) + O}}{{\bf{H}}^{\bf{ - }}}{\bf{(aq) and}}\\{{\bf{C}}_{{\bf{10}}}}{{\bf{H}}_{{\bf{14}}}}{{\bf{N}}_{\bf{2}}}{{\bf{H}}^{\bf{ + }}}{\bf{(aq) + }}{{\bf{H}}_{\bf{2}}}{\bf{O(l)}} \to {{\bf{C}}_{{\bf{10}}}}{{\bf{H}}_{{\bf{14}}}}{{\bf{N}}_{\bf{2}}}{\bf{H}}_{\bf{2}}^{{\bf{2 + }}}{\bf{(aq) + O}}{{\bf{H}}^{\bf{ - }}}{\bf{(aq)}}\end{aligned}\)

We can assume that the concentration change of \({{\bf{C}}_{{\bf{10}}}}{{\bf{H}}_{{\bf{14}}}}{{\bf{N}}_{\bf{2}}}\) in the first reaction and of \({{\bf{C}}_{{\bf{10}}}}{{\bf{H}}_{{\bf{14}}}}{{\bf{N}}_{\bf{2}}}{{\bf{H}}^{\bf{ + }}}\)W in the second reaction is insignificant because the \({{\bf{K}}_{\bf{1}}}\) and \({{\bf{K}}_{\bf{2}}}\) values are so tiny. We may deduce that the concentration of \({{\bf{C}}_{{\bf{10}}}}{{\bf{H}}_{{\bf{14}}}}{{\bf{N}}_{\bf{2}}}\) is \({\bf{0}}{\bf{.050 M}}\) from this.

We can also indicate the concentrations of \({C_{10}}{H_{14}}{N_2}{H^ + }\) and \(O{H^ - }\) as \(x\) and assume that they are equal:

\(\begin{aligned}\;\;\;\;\;\;\;{K_1} &= \frac{{{x^2}}}{{0.050M}}\\7 \times 1{0^{ - 7}} &= \frac{{{x^2}}}{{0.050}}\\\;\;\;\;\;\;\;\;\;x &= \left( {{C_{10}}{H_{14}}{N_2}{H^ + }} \right)\\\;\;\;\;\;\;\;\;\;\;\;\; &= 1.9 \times 1{0^{ - 4}} M.\end{aligned}\)

\(1.9 \times 1{0^{ - 4}} M\)represents the concentration of \({C_{10}}{H_{14}}{N_2}{H^ + }\). Again, we can assume that the concentration change for \({C_{10}}{H_{14}}{N_2}{H^ + }\) in the second reaction is small. We may also assume that the concentrations of \(O{H^ - }\) and \({C_{10}}{H_{14}}{N_2}H_2^{2 + }\) are the same and label them as \(x.\)

\(\begin{aligned}\;\;\;\;\;\;\;\;\;\;\;{K_2} &= \frac{{{x^2}}}{{1.9 \times 1{0^{ - 4}}}}\\1.4 \times 1{0^{ - 11}} &= \frac{{{x^2}}}{{1.9 \times 1{0^{ - 4}}}}\\\;\;\;\;\;\;\;\;\;\;\;\;\;x &= \left( {{C_{10}}{H_{14}}{N_2}H_2^{2 + }} \right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &= 5.2 \times 1{0^{ - 8}} M.\end{aligned}\)

Because the \(O{H^ - }\) is generated in both the first and second reactions, their final concentration is equal to the sum of the first and second reactions:

\(\begin{aligned}\left( {O{H^ - }} \right) &= \left( {1.9 \times 1{0^{ - 4}} + 5.2 \times 1{0^{ - 8}}} \right) M.\\\left( {O{H^ - }} \right) &= 1.9 \times 1{0^{ - 4}} M.\end{aligned}\)

\({C_{10}}{H_{14}}{N_2}\) has a concentration \(0.050M\), \({C_{10}}{H_{14}}{N_2}{H^ + }\) has a concentration of \(1.9 \times 1{0^{ - 4}}M\), \({C_{10}}{H_{14}}{N_2}H_2^{2 + }\) has a concentration of \(5.2 \times 1{0^{ - 8}}M\), and the concentration of \(O{H^ - }\) ions has a concentration of \(1.9 \times 1{0^{ - 4}}M\).

The final solutions are:

\(\begin{aligned}\;\;\;\;\;\;\left( {{C_{10}}{H_{14}}{N_2}} \right) &= 0.050 M.\\\;\;\left( {{C_{10}}{H_{14}}{N_2}{H^ + }} \right) &= 1.9 \times 1{0^{ - 4}} M.\\\left( {{C_{10}}{H_{14}}{N_2}H_2^{2 + }} \right) &= 5.2 \times 1{0^{ - 8}} M.\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {O{H^ - }} \right) &= 1.9 \times 1{0^{ - 4}} M.\end{aligned}\)