# Chapter 13 - Exercises - Page 575d: 83

a) $NH_3(aq) + H_2O(l) \lt -- \gt N{H_4}^+(aq) + OH^-(aq)$ $K_b = \frac{[OH^-][N{H_4}^+]}{[NH_3]}$ b) $C_5H_5N(aq) + H_2O(l) \lt -- \gt C_5H_5NH^+(aq) + OH^-(aq)$ $K_b = \frac{[OH^-][C_5H_5NH^+]}{[C_5H_5N]}$

#### Work Step by Step

1. Write the ionization chemical equation: - Write the reactions where $NH_3$ and $C_5H_5N$ take a proton from a water molecule: $NH_3(aq) + H_2O(l) \lt -- \gt N{H_4}^+(aq) + OH^-(aq)$ $C_5H_5N(aq) + H_2O(l) \lt -- \gt C_5H_5NH^+(aq) + OH^-(aq)$ 2. Now, write the $K_b$ expression: - The $K_b$ expression is the concentrations of the products divided by the concentration of the reactants: $K_b = \frac{[Products]}{[Reactants]}$ a) $K_b = \frac{[OH^-][N{H_4}^+]}{[NH_3]}$ b) $K_b = \frac{[OH^-][C_5H_5NH^+]}{[C_5H_5N]}$ *** We don't consider the $[H_2O]$, because it is the solvent of the solution.

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