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

Question

Carboxylic acids readily dissolve in aqueous sodium bicarbonate, liberating carbon dioxide. Which one of the following statement is correct? (A) The free carboxylic acid and its conjugate base are of comparable stability (B) The free carboxylic acid is more stable than its conjugate base (C) The conjugate base of the carboxylic acid is more stable than the free carboxylic acid (D) The conjugate acid of the carboxylic acid is more stable than the free carboxylic acid

Tardigrade Admin · Accepted Answer

R COOH + NaHCO 3 leftharpoons R COONa + H 2 O + CO 2 or R COOH + HCO 3- leftharpoons R RCO -+ H 2 O + CO 2 conjugate base, R COO -is more stable than the free carboxylic acid. That's why, equilibrium shifts in forward direction. Thus, statement (c) is correct.

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

The free carboxylic acid and its conjugate base are of comparable stability

The free carboxylic acid is more stable than its conjugate base

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

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

$RCOO H + N a H C O_{3} ⇌ RCOON a + H_{2} O + C O_{2}$
or $RCOO H + H C O_{3}^{-} ⇌ RRC O^{-} + H_{2} O + C O_{2}$
conjugate base, $RCO O^{-}$ is more stable than the free carboxylic acid.
That's why, equilibrium shifts in forward direction. Thus, statement (c) is correct.

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

The free carboxylic acid and its conjugate base are of comparable stability

The free carboxylic acid is more stable than its conjugate base

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

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

RCOOH+NaHCO3​⇌RCOONa+H2​O+CO2​ or RCOOH+HCO3−​⇌RRCO−+H2​O+CO2​ conjugate base, RCOO−is more stable than the free carboxylic acid. That's why, equilibrium shifts in forward direction. Thus, statement (c) is correct.

$RCOO H + N a H C O_{3} ⇌ RCOON a + H_{2} O + C O_{2}$
or $RCOO H + H C O_{3}^{-} ⇌ RRC O^{-} + H_{2} O + C O_{2}$
conjugate base, $RCO O^{-}$ is more stable than the free carboxylic acid.
That's why, equilibrium shifts in forward direction. Thus, statement (c) is correct.