Question

Acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\), is a weak organic acid, \(\mathrm{p} K_{\mathrm{a}} 4.76\). Write an equation for the equilibrium reaction of acetic acid with each base. Which equilibria lie considerably toward the left? Which lie considerably toward the right? (a) \(\mathrm{NaHCO}_{3}\) (b) \(\mathrm{NH}_{3}\) (c) \(\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{NaOH}\)

Short answer

Answer: The equilibrium reactions of acetic acid with sodium bicarbonate lie considerably to the left. On the other hand, the equilibrium reactions of acetic acid with water and sodium hydroxide lie considerably to the right.

Step-by-step solution

(Equilibrium Reaction with Sodium Bicarbonate)

For the equilibrium reaction of acetic acid with Sodium Bicarbonate (NaHCO\(_3\)), we have: $$\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{NaHCO}_{3} \longleftrightarrow \mathrm{CH}_{3}\mathrm{COO}^{-} + \mathrm{Na}^{+} + \mathrm{H}_{2}\mathrm{CO}_{3}$$

(Equilibrium Reaction with Ammonia)

For the equilibrium reaction of acetic acid with Ammonia (NH\(_3\)), we have: $$\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{NH}_{3} \longleftrightarrow \mathrm{CH}_{3}\mathrm{COO}^{-} + \mathrm{NH}_{4}^{+}$$

(Equilibrium Reaction with Water)

For the equilibrium reaction of acetic acid with Water (H\(_2\)O), we have: $$\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{H}_{2}\mathrm{O} \longleftrightarrow \mathrm{CH}_{3}\mathrm{COO}^{-} + \mathrm{H}_{3}\mathrm{O}^{+}$$

(Equilibrium Reaction with Sodium Hydroxide)

For the equilibrium reaction of acetic acid with Sodium Hydroxide (NaOH), we have: $$\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{NaOH} \longleftrightarrow \mathrm{CH}_{3}\mathrm{COO}^{-} + \mathrm{H}_{2}\mathrm{O} + \mathrm{Na}^{+}$$

(Adding the \(pK_a\) and \(pK_b\) values)

The relationship between the equilibrium constant, \(K\), and the \(pK\) values is: \(K = 10^{-pK}\). So, the \(K_a\) of the acetic acid is: $$K_a = 10^{-4.76}$$ We have to determine the \(pK_b\) for each base: (a) For NaHCO\(_3\), \(K_b = 10^{-(14 - (\mathrm{p}K_{a1} - \mathrm{p}K_{a2}))}\). (b) For NH\(_3\), \(K_b = 10^{-9.25}\). (c) For water, since it's neutral, \(K_b = 10^{-14}\). (d) For NaOH, user the strong base assumption, \(K_b = 10^{0}\).

(Equilibria that lie considerably to the left or right)

By comparing the acidity and basicity constants, we can determine the direction of the equilibrium: (a) Acetic acid and sodium bicarbonate lie considerably toward the left, as their \(pK\) values are similar. (b) Acetic acid equilibrium with ammonia lies slightly towards the right, as the \(pK_b\) of NH\(_3\) is slightly greater than the \(pK_a\) of acetic acid. (c) The equilibrium of acetic acid with water lies towards the right, as \(pK_a\) of acetic acid is much higher than the \(pK_b\) of water. (d) Acetic acid equilibrium with sodium hydroxide lies considerably toward the right, as the \(pK_b\) of NaOH is much higher than the \(pK_a\) of acetic acid.