Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 16 - Exercises - Page 773: 88


a. $$C{O_3}^{2-}(aq) + H_2O(l) \leftrightharpoons HC{O_3}^-(aq) + OH^-(aq)$$ $$K_b = \frac{[HC{O_3}^-][OH^-]}{[C{O_3}^{2-}]}$$ b. $$C_6H_5NH_2(aq) + H_2O(l) \leftrightharpoons C_6H_5NH{_3}^+(aq) + OH^-(aq)$$ $$K_b = \frac{[C_6H_5NH{_3}^+][OH^-]}{[C_6H_5NH_2]}$$ c. $$C_2H_5NH_2(aq) + H_2O(l) \leftrightharpoons C_2H_5N{H_3}^+(aq) + OH^-(aq)$$ $$K_b = \frac{[C_2H_5N{H_3}^+][OH^-]}{[C_2H_5NH_2]}$$

Work Step by Step

A weak base ionizes, receiving one proton $(H^+)$ from a water molecule, producing its conjugate acid and hydroxide ion. $$K_b = \frac{[Products]}{[Reactants]}$$ * Water does not appear in the Kb expression since it is a liquid.
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