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In the following equation, which is the proton donor and which is the proton acceptor? [tex]\[ CO _3^{2-}(aq) + H_2O(l) \rightarrow HCO_3^{-}(aq) + OH^{-}(aq) \][/tex]
A. Donor: [tex]\(HCO_3^{-}\)[/tex]; acceptor: [tex]\(OH^{-}\)[/tex]
B. Donor: [tex]\(OH^{-}\)[/tex]; acceptor: [tex]\(HCO_3^{-}\)[/tex]
C. Donor: [tex]\(CO_3^{2-}\)[/tex]; acceptor: [tex]\(H_2O\)[/tex]
D. Donor: [tex]\(H_2O\)[/tex]; acceptor: [tex]\(CO_3^{2-}\)[/tex]
we need to understand the basic concept of acid-base reactions involving proton transfers.
1. Identify the proton donor: A proton donor is a species that loses or donates a proton (H⁺) during the reaction. In this context, we need to examine what happens to each species in the reaction.
- Water (H₂O) can donate a proton to become OH⁻. - Assuming water donates a proton, it will transform to OH⁻, which is exactly shown in the product side of the reaction.
2. Identify the proton acceptor: A proton acceptor is a species that gains or accepts a proton during the reaction.
- Carbonate ion (CO₃²⁻) when it accepts a proton, will convert into bicarbonate ion (HCO₃⁻). - Since we see that HCO₃⁻ is a product, it indicates that CO₃²⁻ has gained a proton, thus acting as the proton acceptor.
From this analysis:
- Proton donor: H₂O (which donates a proton and turns into OH⁻) - Proton acceptor: CO₃²⁻ (which accepts the proton and turns into HCO₃⁻)
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