The base dissociation equilibrium constant (Kb) for NH2NH2 is 1.70×10-6 (eq. 1).
(eq. 1): NH2NH2(aq) + H2O(l) = OH-(aq) + NH2NH3+(aq)
How do I calculate the value of the equilibrium constant (Ka) for the hydrolysis of NH2NH3+ as shown in the reaction (eq. 2)?
(eq. 2): NH2NH3+(aq) + H2O(l) = H3O+(aq) + NH2NH2(aq)
Constant is Kw=10-14