Chemistry: Molecular Approach (4th Edition)

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

Chapter 16 - Exercises - Page 771: 42


a. Weak $$K_a = \frac{[F^-][H_3O^+]}{[HF]}$$ b. Weak $$K_a = \frac{[CH{O_2}^-][H_3O^+]}{[HCHO_2]}$$ c. Strong d, Weak $$K_a = \frac{[HC{O_3}^-][H_3O^+]}{[H_2CO_3]}$$

Work Step by Step

If the acid is on the "Strong Acids" table (Table 16.3, page 729), it is a strong acid; if the acid does not appear on this list, it is probably a weak acid. 1. Write the balanced equation for the ionization of this acid: a.$$HNO_3(aq) + H_2O(l) \leftrightharpoons N{O_3}^-(aq) + H_3O^+(aq)$$ b. $$HCHO_2(aq) + H_2O(l) \leftrightharpoons CH{O_2}^-(aq) + H_3O^+(aq)$$ d.$$H_2CO_3(aq) + H_2O(l) \leftrightharpoons HC{O_3}^-(aq) + H_3O^+(aq)$$ 2. The exponent of each concentration is equal to its balance coefficient. a. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ NO{_3}^- ][ H_3O^+ ]}{[ HNO_3 ]}$$ b. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ CHO{_2}^- ][ H_3O^+ ]}{[ HCHO_2 ]}$$ d. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ H{CO_3}^- ][ H_3O^+ ]}{[ H_2CO_3 ]}$$
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