Answer
Reactants:
Acid: $HS{O_4}^-$
Base: $HC{O_3}^-$
Products:
Conjugate base: $S{O_4}^{2-}$
Conjugate acid: $H_2CO_3$
Work Step by Step
Reactants: Left side of the reaction.
Products: Right side.
1. Identify the acid and the base of the reactants:
- $HS{O_4}^-$ is turning into $S{O_4}^{2-}$; therefore, it is donating one proton, so it is the acid.
- $HC{O_3}^-$ is turning into $H_2CO_3$; therefore, it is receiving one proton, so it is the base.
2. Identify the conjugate acid and base:
- The conjugate base is the result of an acid losing a $H^+$. Since $HS{O_4}^-$ is the acid, $S{O_4}^{2-}$ is the conjugate base.
- The conjugate acid is the result of a base gaining a $H^+$. Since $HC{O_3}^-$ is the base, $H_2CO_3$ is the conjugate acid.
