Question

what is the net ionic equation for the reaction of na₂co₃ and hcl in water?
○ 2h⁺(aq) + co₃²⁻(aq) ==> h₂co₃(aq)
○ 2na⁺(aq) + 2cl⁻(aq) ==> na₂cl₂(aq)
○ 2h⁺(aq) + co₃²⁻(aq) ==> co₂(g) + h₂o(l)
○ 2na⁺(aq) + 2cl⁻(aq) ==> 2nacl(aq)

Explanation:

Step1: Recall Reaction of \( \text{Na}_2\text{CO}_3 \) and \( \text{HCl} \)

Sodium carbonate (\( \text{Na}_2\text{CO}_3 \)) and hydrochloric acid (\( \text{HCl} \)) react. First, write the molecular equation: \( \text{Na}_2\text{CO}_3(\text{aq}) + 2\text{HCl}(\text{aq})
ightarrow 2\text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g}) \). Then, write the total ionic equation by dissociating strong electrolytes: \( 2\text{Na}^+(\text{aq}) + \text{CO}_3^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) + 2\text{Cl}^-(\text{aq})
ightarrow 2\text{Na}^+(\text{aq}) + 2\text{Cl}^-(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g}) \).

Step2: Cancel Spectator Ions

Spectator ions are \( \text{Na}^+ \) and \( \text{Cl}^- \) (they appear on both sides). Cancel them to get the net ionic equation: \( 2\text{H}^+(\text{aq}) + \text{CO}_3^{2-}(\text{aq})
ightarrow \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g}) \). Now check the options:

• First option: \( \text{H}_2\text{CO}_3 \) is unstable and decomposes to \( \text{CO}_2 \) and \( \text{H}_2\text{O} \), so this is incorrect.
• Second option: \( \text{Na}_2\text{Cl}_2 \) is not a valid compound, and \( \text{Na}^+ \) and \( \text{Cl}^- \) are spectators, so incorrect.
• Third option: Matches the net ionic equation we derived.
• Fourth option: \( \text{NaCl} \) is soluble, and \( \text{Na}^+ \) and \( \text{Cl}^- \) are spectators, so this is just the formation of \( \text{NaCl} \) which is not the net ionic reaction for the acid - carbonate reaction (the key reaction is between \( \text{H}^+ \) and \( \text{CO}_3^{2-} \) to form \( \text{CO}_2 \) and \( \text{H}_2\text{O} \)), so incorrect.

Answer:

\( 2\text{H}^+_{(aq)} + \text{CO}_3^{2-}_{(aq)} ==> \text{CO}_{2(g)} + \text{H}_2\text{O}_{(l)} \) (the third option)