See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution: Which compound will NOT release $CO_2$ with $NaHCO_3$? **Key Principle:** $NaHCO_3$ is a weak base (pH ≈ 8.3). It releases $CO_2$ only when reacting with acids **stronger than carbonic acid** ($H_2CO_3$, $K_a \approx 4.3 \times 10^{-7}$). **Analysis of each acid:** **(a) Lactic acid:** $K_a \approx 1.4 \times 10^{-4}$ — Much stronger than $H_2CO_3$ → **Releases $CO_2$** ✓ **(b) Tartaric acid:** $K_a \approx 10^{-3}$ — Much stronger than $H_2CO_3$ → **Releases $CO_2$** ✓ **(c) Carbolic acid (Phenol):** $K_a \approx 1 \times 10^{-10}$ — Weaker than $H_2CO_3$ → **Does NOT release $CO_2$** ✗ **(d) Picric acid:** $K_a \approx 0.4$ — Extremely strong acid → **Releases $CO_2$** ✓ **Why (c) fails:** Phenol is too weak an acid. $NaHCO_3$ cannot donate a proton to phenol because $H_2CO_3$ is a stronger acid than phenol. The reaction does not proceed. $$\text{Phenol} + NaHCO_3 \rightarrow \text{No reaction}$$ **Answer: (d) Picric acid** — Wait, re-reading: the question asks which will **NOT** release $CO_2$. **Correct Answer: (c) Carbolic acid (Phenol)** is the compound that will NOT release $CO_2$ gas.