## Chemistry: Atoms First (2nd Edition)

a) $C_6H_5NH_2(aq) + H_2O(l) \lt -- \gt C_6H_5N{H_3}^+(aq) + OH^-(aq)$ $K_b = \frac{[OH^-][C_6H_5N{H_3}^+]}{[C_6H_5NH_2]}$ b) $(CH_3)_2NH(aq) + H_2O(l) \lt -- \gt (CH_3)_2N{H_2}^+(aq) + OH^-(aq)$
1. Write the ionization chemical equation: - Write the reaction where $C_6H_5NH_2$ and $(CH_3)_2NH$ take a proton from a water molecule: a) $C_6H_5NH_2(aq) + H_2O(l) \lt -- \gt C_6H_5N{H_3}^+(aq) + OH^-(aq)$ b) $(CH_3)_2NH(aq) + H_2O(l) \lt -- \gt (CH_3)_2N{H_2}^+(aq) + OH^-(aq)$ 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) $K_b = \frac{[OH^-][C_6H_5N{H_3}^+]}{[C_6H_5NH_2]}$ b) $K_b = \frac{[OH^-][(CH_3)_2N{H_2}^+]}{[(CH_3)_2NH]}$ *** We don't consider the $[H_2O]$, because it is the solvent of the solution.