Question

Are carboxylic acids typically strong acids or weak acids? Write an equation showing the acid \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\) ionizing in water.

Short answer

Carboxylic acids are typically weak acids, meaning they partially dissociate in water. The ionization equation for propanoic acid (CH₃CH₂COOH) in water is: \[ CH_{3}CH_{2}COOH \ + \ H_{2}O \ \rightleftharpoons \ CH_{3}CH_{2}COO^{-} \ + \ H_{3}O^{+} \] This equation shows the acid reacts with water, forming a carboxylate ion (CH₃CH₂COO⁻) and a hydronium ion (H₃O⁺), with the equilibrium indicated by the double arrows (\(\rightleftharpoons\)).

Step-by-step solution

Identify the strength of carboxylic acids

Carboxylic acids are typically weak acids. They partially dissociate in water, producing a relatively small concentration of hydrogen ions (H⁺) and leaving the majority of the acid molecules intact.

Write the ionization equation for CH₃CH₂COOH in water

To show the ionization of CH₃CH₂COOH (propanoic acid) in water, we need to write a balanced equation. The reaction involves the transfer of a proton (H⁺) from the carboxylic acid to a water molecule, forming a hydronium ion (H₃O⁺) and a carboxylate ion. The ionization equation for propanoic acid (CH₃CH₂COOH) in water is: \[ CH_{3}CH_{2}COOH \ + \ H_{2}O \ \rightleftharpoons \ CH_{3}CH_{2}COO^{-} \ + \ H_{3}O^{+} \] This equation shows that propanoic acid (CH₃CH₂COOH) reacts with water (H₂O) and reaches an equilibrium with the conjugate base CH₃CH₂COO⁻ and the hydronium ion H₃O⁺. The double arrows (\(\rightleftharpoons\)) indicate a state of equilibrium, where the forward and reverse reactions occur at the same rate, typical of weak acids in water.