## Chemistry and Chemical Reactivity (9th Edition)

Since Kb1 is much bigger than Kb2, we can simplify the problem by assuming that the dissociation of $N_2H_5^+$ and formation of $OH^-$ in the second reaction won't affect the first equilibrium. 1) $K_{b1}=[N_2H_5^+][OH^-]/[N_2H_4]$ $8.5\times10^{-7}=x\dot{}x/(0.025-x)$ $x=1.45\times10^{-4}\ M$ 2) $K_{b2}=[N_2H_6^{2+}][OH^-]/[N_2H_5^+]$ $8.9\times10^{-16}=y(1.45\times10^{-4}+y)/(1.45\times10^{-4}-y)$ $y\approx8.9\times10^{-16}\ M$ $[OH^-]\approx1.45\times10^{-4}\ M$ $pH=14+log([OH^-])=10.16$