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Carboxylic acids readily dissolve in aqueous sodium bicarbonate, liberating carbon dioxide. Which one of the following is correct?

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Free carboxylic acid and its conjugate base are of comparable stability

The conjugate base of the carboxylic acid is more stable than the free carboxylic acid

The free carboxylic acid is more stable than its conjugate base

The conjugate acid of the carboxylic acid is more stable than the free carboxylic acid

answer is C.

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Detailed Solution

$RCOOH + {NaHCO}_{3} ⇌ RCOONa + H_{2} O + {CO}_{2}$

or $RCOOH + {HCO}_{3}^{-} ⇌ {RCOO}^{-} + H_{2} O + {CO}_{2}$

conjugate base, ${RCOO}^{-}$ is more stable. That is why equilibrium shifts in forward direction.

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