## Chemistry 10th Edition

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