Answer
${HPO_4}^{2-} + H_2O -> {H_2PO_4}^{-} + OH^-$
$Kb = \frac{[{H_2PO_4}^-][OH^-]}{[{HPO_4}^{2-}]}$
Work Step by Step
1. Write the reaction:
${HPO_4}^{2-} + H_2O -> {H_2PO_4}^{-} + OH^-$
2. Now, write the Kb formula:
$Kb = \frac{[Conjugate.Acid]*[OH^-]}{[Base]}$
$Kb = \frac{[{H_2PO_4}^-][OH^-]}{[{HPO_4}^{2-}]}$