Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 16 Principles of Chemical Reactivity: The Chemistry of Acids and Bases - Study Questions - Page 629d: 73


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Work Step by Step

First equilibrium: $K_{b1}=[N_2H_5^+][OH^-]/[N_2H_4]$ $8.5\cdot10^{-7}=x^2/(0.010\ M-x)$ $x=9.18\cdot10^{-5}\ M$ Second equilibrium: $K_{b2}=[N_2H_6^{2+}][OH^-]/[N_2H_5^+]$ $8.9\cdot10^{-16}=y(9.18\cdot10^{-5}+y)/(9.18\cdot10^{-5}-y)$ $y=8.9\cdot10^{-16}\ M$ a) $[N_2H_6^{2+}]=8.9\cdot10^{-16}\ M, [OH^-]=[N_2H_5^+]=9.18\cdot10^{-5}\ M$ b) $pH=14+log(9.18\cdot10^{-5})=9.96$
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