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]



Answer :

To determine the proton donor and acceptor in the reaction:

[tex]\[ CO_3^{2-}(aq) + H_2O(l) \rightarrow HCO_3^-(aq) + OH^-(aq) \][/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⁻)

Therefore, the correct identification is:

- Donor: [tex]\( H_2O \)[/tex]
- Acceptor: [tex]\( CO_3^{2-} \)[/tex]

Thus, the correct answer is:

- Donor: [tex]\( H_2O \)[/tex]; acceptor: [tex]\( CO_3^{2-} \)[/tex]