Answer
${CHO_2}^- + H_2O -> HCOOH + OH^- $
$Kb = \frac{[HCOOH][OH^-]}{{CHO_2}^-}$
Work Step by Step
1. Write and balance the reaction with water:
${CHO_2}^- + H_2O -> HCOOH + OH^- $
$1{CHO_2}^- + 1H_2O -> 1HCOOH + 1OH^- $
2. Now, write the Kb formula:
$Kb = \frac{[Conjugate.acid]*[OH^-]}{[Base]}$
$Kb = \frac{[HCOOH][OH^-]}{{CHO_2}^-}$
