Chapter 16 Principles of Chemical Reactivity: The Chemistry of Acids and Bases - Study Questions - Page 629d: 74

See the answer below.

First equilibrium: $K_{b1}=[BH^+][OH^-]/[B]$ $8.5\cdot10^{-5}=x^2/(0.15\ M-x)$ $x=0.0035\ M$ Second equilibrium: $K_{b2}=[BH_2^{2+}][OH^-]/[BH^+]$ $8.9\cdot10^{-16}=y(0.0035+y)/(0.0035-y)$ $y=8.9\cdot10^{-16}\ M$ $[BH_2^{2+}]=8.9\cdot10^{-16}\ M, [OH^-]=0.0035\ M$