Consider nicotine,
and its basic equilibriums,
#C_10H_14N_2(aq) + H_2O(l) rightleftharpoons HC_10H_14N_2^(+)(aq) + OH^(-)(aq)#
where #K_("b"_1)= ([HC_10H_14N_2^+][OH^-])/([C_10H_14N_2]) approx 1.0*10^-6#, and
#HC_10H_14N_2^(+)(aq) +H_2O(l) rightleftharpoons H_2C_10H_14N_2^(2+)(aq) + OH^(-)(aq)#
where #K_("b"_2) = ([H_2C_10H_14N_2^(2+)][OH^-])/([HC_10H_14N_2^(+)]) approx 1.3*10^-11#
Let's consider each abstraction at a time,
#K_("b"_1) = x^2/(1.15*10^-3"M") = 1.0*10^-6#
#therefore x_1 = [OH^-]=[HC_10H_14N_2^+] approx 3.4*10^-5"M"#, and
#K_("b"_2) = x^2/(3.4*10^-5"M"-x) = 1.3*10^-11#
#therefore x_2 = [OH^-] approx 2.1*10^-8"M"#
To solve for pH, we need to find the pOH and derive it from there,
#"pOH" = -log(x_1+x_2) approx 4.47#
#therefore "pH" = 14 - "pOH" approx 9.53#
given your data.
If you care: each nitrogen on nicotine is weakly basic and will collect protons at a rate described by the #K_"b"# values. If I was to guess, the pyridine analog on the left would pick up a proton first, followed by the other nitrogen. This is the stuff you start to think about when you begin organic chemistry.
