Respuesta :
Acetic Acid + Sodium ethoxide ⇄ Butyric Acid + Sodium Hydroxide
Explanation:
An ionic equation for the reaction of acetic acid with sodium ethoxide is as follows -
- Reactants -
Acetic Acid and Sodium ethanolate (sodium ethoxide)
- Products -
Butyric Acid and Sodium hydroxide
Hence,
Acetic Acid + Sodium ethoxide ⇄ Butyric Acid + Sodium Hydroxide
[tex]CH_3COOH + C_2H_5ONa[/tex] ⇄ [tex]CH_3COOC_2H_5 + NaOH[/tex]
- Weak acids and bases have low energy than strong acids and bases.
- The chemical equilibria shift the reaction side with the species having lower energy.
- Given reaction is an acid-base reaction in which the equilibrium favors the starting material that means it will go to the side of the weakest acid that is acetic acid is weaker than butyric acid.
The ionic equation for the reaction of acetic acid with sodium ethoxide is:
[tex]\mathbf{CH_3COOH _{(aq)} + CH_3CH_2O^-_{(aq)} \to CH_3COO^-_{(aq)} + CH_3CH_2OH_{(aq)}}[/tex]
The acid dissociation constant at equilibrium is greater than 1, the equilibrium of the reaction will favor the products.
Ionic equations are chemical reactions expressed in a way in which all the species participating in the reaction exist in their free state and the ions that are absent in the overall chemical reaction are known to be spectators ions.
The ionic equation required can be expressed as:
[tex]\mathbf{CH_3COOH _{(aq)} + CH_3CH_2O^-_{(aq)} \to CH_3COO^-_{(aq)} + CH_3CH_2OH_{(aq)}}[/tex]
stronger acid stronger base conjugate base conjugate acid pKa = 16
with a pKa = 4.7
At equilibrium, the pKa of the reaction is:
= 4.7 - 16
= 11.3
The acid dissociation constant Ka at equilibrium is:
[tex]\mathbf{K_a = 10^{-pKa}}[/tex]
[tex]\mathbf{K_a = 10^{-(-11.3)}}[/tex]
[tex]\mathbf{K_a = 10^{11.3}}[/tex]
Therefore, since the acid dissociation constant at equilibrium is greater than 1, the equilibrium of the reaction will favor the products.
Learn more about acid dissociation constant here:
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