Answer
Kb ($CH_3COO^-$) = $5.56 \times 10^{-10}$
Kb ($ClO^-$) = $3.33 \times 10^{-7}$
Work Step by Step
For conjugate acid-base pairs we got:
$Ka * Kb = Kw = 10^{-14}$
- ($CH_3COO^-$) is the conjugate base of acetic acid ($Ka = 1.8 \times 10^{-5}$), therefore:
$1.8 \times 10^{-5} * Kb = 10^{-14}$
$Kb = \frac{10^{-14}}{1.8 \times 10^{-5}}$
$Kb = 5.56 \times 10^{-10}$
- $ClO^-$ is the conjugate base of hypochlorous acid ($Ka = 3 \times 10^{-8} $), therefore:
$3 \times 10^{-8} * Kb = 10^{-14}$
$Kb = 3.33 \times 10^{-7}$
