RCOOH + NaHCO_3 to RCOONa + H_2O + CO_2. In t | vedclass

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(B) The reaction between a carboxylic acid $(RCOOH)$ and sodium bicarbonate $(NaHCO_3)$ is an acid-base reaction.$RCOOH + NaHCO_3 	o RCOONa + H_2O +

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$NaHCO_3$

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(B) The reaction between a carboxylic acid $(RCOOH)$ and sodium bicarbonate $(NaHCO_3)$ is an acid-base reaction.$RCOOH + NaHCO_3 	o RCOONa + H_2O + CO_2$In this process,the $H^+$ ion from the carboxylic acid reacts with the $HCO_3^-$ ion from sodium bicarbonate to form carbonic acid $(H_2CO_3)$,which is unstable and decomposes into water $(H_2O)$ and carbon dioxide $(CO_2)$.Since the $CO_2$ is produced from the decomposition of the bicarbonate ion $(HCO_3^-)$,the evolved $CO_2$ comes from $NaHCO_3$.

Column-$1$ (Acid)	Column-$2$ ($pK_a$ value)
$A. CH_3COOH$	$I. 0.23$
$B. F_3CCOOH$	$II. 3.41$
$C. \text{Benzoic acid}$	$III. 4.19$
$D. p\text{-Nitrobenzoic acid}$	$IV. 4.76$

$RCOOH + NaHCO_3 \to RCOONa + H_2O + CO_2$. In this reaction,the $CO_2$ evolved comes from:

Similar Questions

Match the following:
Column-$1$ (Acid) Column-$2$ ($pK_a$ value)
$A. CH_3COOH$ $I. 0.23$
$B. F_3CCOOH$ $II. 3.41$
$C. \text{Benzoic acid}$ $III. 4.19$
$D. p\text{-Nitrobenzoic acid}$ $IV. 4.76$

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