1. In general: $$\ce{acid + H2O <=> base + H_3O^+}$$
2. It is called the Henderson-Hasselbalch equation $$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\ce{[base]}}{\ce{[acid]}}$$
3. I believe the equilibrium constant for a such a neutralisation would look like $$K_n = K_\mathrm{a}K_\mathrm{b}\frac{1}{K_\mathrm{w}}$$