Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 14 - Acids and Bases - Exercises: 36

Answer

a. $$K_a=\frac{[H^+][CN^-]}{[HCN]}$$ b. $$K_a=\frac{[H^+][OC_6H_5^-]}{[HOC_6H_5]}$$ c. $$K_a=\frac{[H^+][C_6H_5NH_2]}{[C_6H_5NH_3]}$$

Work Step by Step

The general formula for the $K_a$ equilibrium constant for the reaction: $$HA\Rightarrow H^++A^-$$ is $K_a=\frac{[H^+][A^-]}{[HA]}$. Therefore, the dissociation reactions are as follows: $$HCN_{(aq)}\Rightarrow H^+_{(aq)}+CN^-_{(aq)}$$ $$HOC_6H_{5(aq)}\Rightarrow H^+_{(aq)}+OC_6H_{5(aq)}^-$$ $$C_6H_5NH_{3(aq)}\Rightarrow H^+_{(aq)}+C_6H_5NH_{2(aq)}$$ and the dissociation constants are: $$K_a=\frac{[H^+][CN^-]}{[HCN]}$$ $$K_a=\frac{[H^+][OC_6H_5^-]}{[HOC_6H_5]}$$ $$K_a=\frac{[H^+][C_6H_5NH_2]}{[C_6H_5NH_3]}$$
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