Nicotine is responsible for the addictive properties of tobacco. assume kb1 = 1.0×10-6 and kb2 = 1.3×10-11 . determine the ph of a 1.85×10-3 m solution of nicotine.]
If we abbreviate the formula for nicotine as Nic, then the equations for two different equilibria of Nic in water are Nic + H2O ---> NicH+ + OH- NicH+ + H2O ---> NicH2 2+ + OH-
We can write the Kb1 expression for the first equation as Kb1 = 1.0×10^-6 = [NicH+][OH-] / [Nic] 1.0×10^-6 = x^2 / 1.85×10^-3 - x Approximating that x is negligible compared to 1.85×10^-3 simplifies the equation to 1.0×10^-6 = x^2 / 1.85×10^-3 x = 0.0000430 x = [OH-] = 4.30×10^-5 M
From the Kb2 expression Kb2 = 1.3×10-11 = [NicH2 2+][OH-] / [NicH+] 1.1×10^-10 = x^2 / 4.30×10^-5 - x Approximating that x is negligible compared to 4.30×10^-5 simplifies the equation to 1.1×10^-10 = x^2 / 4.30×10^-5 x = [OH-] = 6.88×10^-8
The concentration [OH-] can be computed as [OH-] = 4.30×10^-5 M + 6.88×10^-8 M = 4.30×10^-5 M This shows that the second equilibrium has a negligible effect on the pH.
We can now calculate for pH: pOH = -log [OH-] = -log (4.30×10^-5 M) = 4.37 pH = 14 - pOH = 14 - 4.37 = 9.63
