Chemistry

Can somone please explain this problem to me? The hydrolysis equilibrium constant (Ka) for NH2NH3+ is 5.88E-9 (eq. 1).

(eq. 1): NH2NH3+(aq) + H2O(l) = H3O+(aq) + NH2NH2(aq)

Calculate the value of the base dissociation constant (Kb) for NH2NH2 as shown in the reaction (eq. 2).

(eq. 2): NH2NH2(aq) + H2O(l) = OH-(aq) + NH2NH3+(aq)

I solved it but I was one decimal place off. I am double checking but I cannot find my source of error. My answer was 1.7006E-5 by the way.

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  1. Kb = Kw/Ka = 1 x 10^-14/5.88 x 10^-9 = 1.7 x 10^-6.

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