Chemistry: Molecular Approach (4th Edition)

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

Chapter 16 - Exercises - Page 771: 40

Answer

$HC{O_3}^-$ is acting as an acid and $HS^-$ is acting as a base: $$HC{O_3}^-(aq) + HS^-(aq) \leftrightharpoons C{O_3}^{2-}(aq) + H_2S(aq)$$ $HS^-$ is acting as an acid and $HC{O_3}^-$ is acting as a base: $$HC{O_3}^-(aq) + HS^-(aq) \leftrightharpoons H_2C{O_3}(aq) + S^{2-}(aq)$$

Work Step by Step

1. Write the first equation. Since $HC{O_3}^-$ is acting as an acid, it must donate one proton to the other molecule, and since $HS^-$ is acting as a base, it must accept one proton from the other substance. 2. For the second one, we just invert the roles. $HC{O_3}^-$ will receive one proton and $HS^-$ will donate one.
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