Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning
ISBN 10: 1305079248
ISBN 13: 978-1-30507-924-3

Chapter 13 - Exercises - Page 575e: 106


$H_3C_6H_5O_7(aq) + H_2O(l) \lt -- \gt H_2C_6H_5{O_7}^-(aq) + H_3O^+(aq)$ $H_2C_6H_5{O_7}^-(aq) + H_2O(l) \lt -- \gt HC_6H_5{O_7}^{2-}(aq) + H_3O^+(aq)$ $HC_6H_5{O_7}^{2-}(aq) + H_2O(l) \lt -- \gt C_6H_5{O_7}^{3-}(aq) + H_3O^+(aq)$

Work Step by Step

1. Write the reaction where $H_3C_6H_5O_7$ behaves as an acid in water. $H_3C_6H_5O_7(aq) + H_2O(l) \lt -- \gt H_2C_6H_5{O_7}^-(aq) + H_3O^+(aq)$ - It will donate one proton to the water molecule, producing $H_3O^+$. 2. Write the reaction where the conjugate base acts as an acid: $H_2C_6H_5{O_7}^-(aq) + H_2O(l) \lt -- \gt HC_6H_5{O_7}^{2-}(aq) + H_3O^+(aq)$ - Repeat step 2 until the conjugate base is incapable of donating protons. $HC_6H_5{O_7}^{2-}(aq) + H_2O(l) \lt -- \gt C_6H_5{O_7}^{3-}(aq) + H_3O^+(aq)$
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