Answer
$K_b ({N_3}^-)= 5.3\times 10^{- 10}$
Work Step by Step
1. Since ${N_3}^-$ is the conjugate base of $HN_3$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 1.0 \times 10^{-14}$
$ 1.9\times 10^{- 5} * K_b = 1.0 \times 10^{-14}$
$K_b = \frac{1.0 \times 10^{-14}}{ 1.9\times 10^{- 5}}$
$K_b = 5.3\times 10^{- 10}$
