## Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning

# Chapter 14 - Acids and Bases - Questions for Review and Thought - Topical Questions - Page 652c: 43c

#### Answer

$S{O_3}^{2-}(aq) + H_2O(l) \lt -- \gt HS{O_3}^-(aq) + OH^-(aq)$ $K_b = \frac{[OH^-][HS{O_3}^-]}{[S{O_3}^{2-}]}$

#### Work Step by Step

1. Write the ionization chemical equation: - Since $S{O_3}^{2-}$ is a base, write the reaction where it takes a proton from a water molecule: $S{O_3}^{2-}(aq) + H_2O(l) \lt -- \gt HS{O_3}^-(aq) + OH^-(aq)$ 2. Now, write the $K_b$ expression: - The $K_b$ expression is the concentrations of the products divided by the concentration of the reactants: $K_b = \frac{[Products]}{[Reactants]}$ $K_b = \frac{[OH^-][HS{O_3}^-]}{[S{O_3}^{2-}]}$ *** We don't consider the $[H_2O]$, because it is the solvent of the solution.

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