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15240672380's Question
Posted 3 months ago

Consider the following Brønsted - Lowry equilibrium system:
\mathrm{HSO}_{3}^{-}+\mathrm{H}_{2} \mathrm{PO}_{4}^{-} \rightleftharpoons \mathrm{SO}_{3}^{2-}+\mathrm{H}_{3} \mathrm{PO}_{4}

What are the two Brønsted - Lowry bases in the equilibrium above?
H2PO4and SO32HSO3and H3PO4H2PO4and H3PO4HSO3and SO32
\mathrm{H}_{2} \mathrm{PO}_{4}^{-} \text {and } \mathrm{SO}_{3}^{2-} \\
\mathrm{HSO}_{3}^{-} \text {and } \mathrm{H}_{3} \mathrm{PO}_{4} \\
\mathrm{H}_{2} \mathrm{PO}_{4}^{-} \text {and } \mathrm{H}_{3} \mathrm{PO}_{4} \\
\mathrm{HSO}_{3}^{-} \text {and } \mathrm{SO}_{3}^{2-}
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Answer from Sia
Posted 3 months ago
Key Concept
Brønsted-Lowry acid-base theory
In the Brønsted-Lowry acid-base theory, an acid is a proton (H⁺) donor, and a base is a proton acceptor. In the given equilibrium, HSO3\mathrm{HSO}_{3}^{-} acts as an acid by donating a proton to become SO32\mathrm{SO}_{3}^{2-}, and H3PO4\mathrm{H}_{3}\mathrm{PO}_{4} acts as an acid by donating a proton to become H2PO4\mathrm{H}_{2}\mathrm{PO}_{4}^{-}. Therefore, the corresponding bases that accept the protons are SO32\mathrm{SO}_{3}^{2-} and H2PO4\mathrm{H}_{2}\mathrm{PO}_{4}^{-}. However, since the question asks for the bases in the equilibrium, the correct answer is the species that are acting as bases, which are HSO3\mathrm{HSO}_{3}^{-} as it can accept a proton to revert to its acid form, and H3PO4\mathrm{H}_{3}\mathrm{PO}_{4} as it can donate a proton to revert to its conjugate base form, H2PO4\mathrm{H}_{2}\mathrm{PO}_{4}^{-}.

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