Diprotic Acids and Bases: Acid Dissociation and Equilibria

Diprotic Acids and Bases

Acid Dissociation Constant

Diprotic acids are a class of acids capable of donating two acidic hydrogens in aqueous solution. Their dissociation occurs in two distinct steps, each characterized by its own acid dissociation constant (Ka).

• Diprotic Acids (H2A): Can donate two protons (H+), resulting in two Ka values.

• First dissociation: Removal of the first proton (Ka1).

• Second dissociation: Removal of the second proton (Ka2).

• The relationship between the Ka values and their respective equilibria is important for understanding acid strength and behavior in solution.

Diprotic Acid Equilibria

The dissociation of a diprotic acid can be represented as follows:

Species	Equilibrium	Ka Expression
H2A (aq)	H2A → H+ + HA-	$K_{a1} = \frac{[H^+][HA^-]}{[H_2A]}$
HA- (aq)	HA- → H+ + A2-	$K_{a2} = \frac{[H^+][A^{2-}]}{[HA^-]}$

Additional info: The first dissociation constant (Ka1) is typically larger than the second (Ka2), indicating that the first proton is more easily removed than the second.

Example: Carbonic Acid Dissociation

Carbonic acid (H2CO3) is a weak diprotic acid with the following dissociation constants:

• $K_{a1} = 4.3 \times 10^{-7}$

• $K_{a2} = 5.6 \times 10^{-11}$

To determine the base dissociation constant associated with the carbonate ion (CO32-), use the relationship between acid and base dissociation constants:

• $K_b = \frac{K_w}{K_a}$, where $K_w$ is the ion-product constant for water ($1.0 \times 10^{-14}$ at 25°C).

Example Calculation: For CO32-, the base dissociation constant is:

• $K_b = \frac{K_w}{K_{a2}}$

• Substitute values: $K_b = \frac{1.0 \times 10^{-14}}{5.6 \times 10^{-11}} = 1.8 \times 10^{-4}$

Practice Problem

Determine the equilibrium expression for the Ka of hydrofluoric acid (HF):

• $K_a = \frac{[H^+][F^-]}{[HF]}$