Answer
$K_b = 8.913\times 10^{- 11}$
Work Step by Step
1. Calculate the Ka value:
$K_a = 10^{-pKa}$
$K_a = 10^{- 3.95}$
$K_a = 1.122 \times 10^{- 4}$
2. Since $[Cr{(H_2O)_5OH}]^{2+}$ is the conjugate base of $[Cr{(H_2O)_6}]^{3+}$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 1.122\times 10^{- 4} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 1.122\times 10^{- 4}}$
$K_b = 8.913\times 10^{- 11}$
