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

In summary, the equilibrium reactions for acetic acid with each base are as follows: (a) With NaHCO3: $$\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{HCO}_{3}^{-} \rightleftharpoons \mathrm{CH}_{3}\mathrm{COO}^{-} + \mathrm{H}_{2}\mathrm{CO}_{3}$$ The equilibrium lies to the left, favoring the formation of acetic acid and bicarbonate ion. (b) With NH3: $$\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{NH}_{3} \rightleftharpoons \mathrm{CH}_{3}\mathrm{COO}^{-} + \mathrm{NH}_{4}^{+}$$ The equilibrium lies to the right, favoring the formation of the acetate ion and ammonium ion. (c) With H2O: $$\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{CH}_{3}\mathrm{COO}^{-} + \mathrm{H}_{3}\mathrm{O}^{+}$$ The equilibrium lies to the right, favoring the formation of the acetate ion and hydronium ion. (d) With NaOH: $$\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{OH}^{-} \rightleftharpoons \mathrm{CH}_{3}\mathrm{COO}^{-} + \mathrm{H}_{2}\mathrm{O}$$ The equilibrium lies far to the right, favoring the formation of the acetate ion and water.

Step-by-step solution

(a) Equilibrium reaction with NaHCO3

To write the equilibrium reaction of acetic acid with NaHCO3, first identify the relevant species: the acetic acid molecule, \(\mathrm{CH}_{3} \mathrm{COOH}\), and the bicarbonate ion, \(\mathrm{HCO}_{3}^{-}\). The equilibrium reaction is: $$\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{HCO}_{3}^{-} \rightleftharpoons \mathrm{CH}_{3}\mathrm{COO}^{-} + \mathrm{H}_{2}\mathrm{CO}_{3}$$

(a1) Determining the direction of the equilibrium

To determine the direction of the equilibrium, find the sum of the pKa values for acetic acid and carbonic acid (which can be found from the equilibrium constant of the \(\mathrm{H}_{2}\mathrm{CO}_{3}\) dissociation). If the sum is less than the pKa of the conjugate acid, the equilibrium will lie to the right (formation of the weaker acid, \(\mathrm{H}_{2}\mathrm{CO}_{3}\)). If the sum is greater than the pKa of the conjugate acid, the equilibrium will lie to the left (formation of the stronger acid, \(\mathrm{CH}_{3}\mathrm{COOH}\)).

(b) Equilibrium reaction with NH3

To write the equilibrium reaction of acetic acid with NH3, first identify the relevant species: the acetic acid molecule, \(\mathrm{CH}_{3} \mathrm{COOH}\), and the ammonia molecule, \(\mathrm{NH}_{3}\). The equilibrium reaction is: $$\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{NH}_{3} \rightleftharpoons \mathrm{CH}_{3}\mathrm{COO}^{-} + \mathrm{NH}_{4}^{+}$$

(b1) Determining the direction of the equilibrium

To determine the direction of the equilibrium, compare the pKa value of acetic acid to that of the ammonium ion, \(\mathrm{NH}_{4}^{+}\). If the pKa of acetic acid is greater than the pKa of the ammonium ion, the equilibrium will lie to the left (formation of the stronger acid, \(\mathrm{CH}_{3}\mathrm{COOH}\)). If the pKa of acetic acid is less than the pKa of the ammonium ion, the equilibrium will lie to the right (formation of the weaker acid, \(\mathrm{NH}_{4}^{+}\)).

(c) Equilibrium reaction with H2O

To write the equilibrium reaction of acetic acid with H2O, first identify the relevant species: the acetic acid molecule, \(\mathrm{CH}_{3} \mathrm{COOH}\), and the water molecule, \(\mathrm{H}_{2}\mathrm{O}\). The equilibrium reaction is: $$\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{CH}_{3}\mathrm{COO}^{-} + \mathrm{H}_{3}\mathrm{O}^{+}$$

(c1) Determining the direction of the equilibrium

The pKa value given for acetic acid is 4.76, and the pKa for the hydronium ion, \(\mathrm{H}_{3}\mathrm{O}^{+}\), is 0. In this case, since the pKa of acetic acid is greater than the pKa of hydronium, the equilibrium will lie to the right (formation of the weaker acid, \(\mathrm{H}_{3}\mathrm{O}^{+}\)).

(d) Equilibrium reaction with NaOH

To write the equilibrium reaction of acetic acid with NaOH, first identify the relevant species: the acetic acid molecule, \(\mathrm{CH}_{3} \mathrm{COOH}\), and the hydroxide ion, \(\mathrm{OH}^{-}\). The equilibrium reaction is: $$\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{OH}^{-} \rightleftharpoons \mathrm{CH}_{3}\mathrm{COO}^{-} + \mathrm{H}_{2}\mathrm{O}$$

(d1) Determining the direction of the equilibrium

Since hydroxide is a strong base, it will react completely with the acetic acid to form water and the acetate ion, \(\mathrm{CH}_{3}\mathrm{COO}^{-}\). In this case, the equilibrium will lie far to the right (formation of the weaker acid, \(\mathrm{H}_{2}\mathrm{O}\)).