Answer
$Ka : N{H_4}^+ = 5.6 \times 10^{-10}$
$Ka : H_3NOH^+ = 9.1 \times 10^{-7}$
Work Step by Step
$K_a * K_b = K_w$
*Just if the acid and the base are conjugate pairs.
$N{H_4}^+:$
$K_a * 1.8 * 10^{-5} = 10^{-14} $
$K_a = \frac{10^{-14}}{1.8 * 10^{-5}}$
$K_a = 5.6 \times 10^{-10}$
$H_3NOH^+:$
$K_a * 1.1 * 10^{-8} = 10^{-14} $
$K_a = \frac{10^{-14}}{1.1 * 10^{-8}}$
$K_a = 9.1 \times 10^{-7}$
