Compound(s) which will liberate carbon dioxid | vedclass

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(C) Sodium bicarbonate $(NaHCO_3)$ is a weak base. It reacts with acids that are stronger than carbonic acid ($H_2CO_3$,$pK_a \approx 6.35$) to libera

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$B$ and $C$ only

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(C) Sodium bicarbonate $(NaHCO_3)$ is a weak base. It reacts with acids that are stronger than carbonic acid ($H_2CO_3$,$pK_a \approx 6.35$) to liberate $CO_2$ gas.$1$. $B$ (Benzoic acid,$pK_a \approx 4.2$) is a stronger acid than $H_2CO_3$,so it reacts with $NaHCO_3$ to liberate $CO_2$.$2$. $C$ (Picric acid,$pK_a \approx 0.38$) is a very strong acid due to the electron-withdrawing effect of three $-NO_2$ groups. It is much stronger than $H_2CO_3$ and thus liberates $CO_2$ with $NaHCO_3$.$3$. $A$ ($2$,$4$,$6$-triamino phenol) is a very weak acid because the $-NH_2$ groups are electron-donating by resonance,which destabilizes the phenoxide ion. It is weaker than $H_2CO_3$ and does not liberate $CO_2$ with $NaHCO_3$.Therefore,both $B$ and $C$ will liberate $CO_2$.

Compound$(s)$ which will liberate carbon dioxide with sodium bicarbonate solution is/are:
$A = \text{2,4,6-triamino phenol}$
$B = \text{Benzoic acid}$
$C = \text{Picric acid (2,4,6-trinitrophenol)}$

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Compound$(s)$ which will liberate carbon dioxide with sodium bicarbonate solution is/are:$A = \text{2,4,6-triamino phenol}$$B = \text{Benzoic acid}$$C = \text{Picric acid (2,4,6-trinitrophenol)}$