Question

Write out the stepwise \(K_{\mathrm{a}}\) reactions for citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\), a triprotic acid.

Short answer

The stepwise $K_{a}$ reactions for citric acid, a triprotic acid, are as follows: 1. First dissociation reaction: \[ H_{3}C_{6}H_{5}O_{7}(aq) \rightleftharpoons H_{2}C_{6}H_{5}O_{7}^{-}(aq) + H_{3}O^{+}(aq) \] with $K_{a1} = \frac{[H_{2}C_{6}H_{5}O_{7}^{-}][H_{3}O^{+}]}{[H_{3}C_{6}H_{5}O_{7}]}$ 2. Second dissociation reaction: \[ H_{2}C_{6}H_{5}O_{7}^{-}(aq) \rightleftharpoons HC_{6}H_{5}O_{7}^{2-}(aq) + H_{3}O^{+}(aq) \] with $K_{a2} = \frac{[HC_{6}H_{5}O_{7}^{2-}][H_{3}O^{+}]}{[H_{2}C_{6}H_{5}O_{7}^{-}]}$ 3. Third dissociation reaction: \[ HC_{6}H_{5}O_{7}^{2-}(aq) \rightleftharpoons C_{6}H_{5}O_{7}^{3-}(aq) + H_{3}O^{+}(aq) \] with $K_{a3} = \frac{[C_{6}H_{5}O_{7}^{3-}][H_{3}O^{+}]}{[HC_{6}H_{5}O_{7}^{2-}]}$

Step-by-step solution

Write the first dissociation reaction

The first proton is removed from citric acid, forming the first conjugate base and releasing a hydronium ion: \[ H_{3}C_{6}H_{5}O_{7}(aq) \rightleftharpoons H_{2}C_{6}H_{5}O_{7}^{-}(aq) + H_{3}O^{+}(aq) \]

Write the second dissociation reaction

The second proton is removed from the first conjugate base, forming the second conjugate base and releasing another hydronium ion: \[ H_{2}C_{6}H_{5}O_{7}^{-}(aq) \rightleftharpoons HC_{6}H_{5}O_{7}^{2-}(aq) + H_{3}O^{+}(aq) \]

Write the third dissociation reaction

The third proton is removed from the second conjugate base, forming the third conjugate base and releasing another hydronium ion: \[ HC_{6}H_{5}O_{7}^{2-}(aq) \rightleftharpoons C_{6}H_{5}O_{7}^{3-}(aq) + H_{3}O^{+}(aq) \]

Write the Ka expressions for each step

The Ka expressions for each step are as follows: For the first dissociation step: \[ K_{a1} = \frac{[H_{2}C_{6}H_{5}O_{7}^{-}][H_{3}O^{+}]}{[H_{3}C_{6}H_{5}O_{7}]} \] For the second dissociation step: \[ K_{a2} = \frac{[HC_{6}H_{5}O_{7}^{2-}][H_{3}O^{+}]}{[H_{2}C_{6}H_{5}O_{7}^{-}]} \] For the third dissociation step: \[ K_{a3} = \frac{[C_{6}H_{5}O_{7}^{3-}][H_{3}O^{+}]}{[HC_{6}H_{5}O_{7}^{2-}]} \]

Key concepts

Triprotic Acid

A triprotic acid is a type of acid that has three ionizable hydrogen atoms, which means it can release three protons (H⁺ ions) into a solution. Citric acid is a classic example of a triprotic acid. In aqueous solutions, each of these hydrogen atoms can dissociate one by one. This property allows triprotic acids to undergo three separate dissociation reactions.
One after the other, each hydrogen gets released as an H⁺ ion, forming a conjugate base at each step. This sequential release results in three unique conjugate bases, each with varying levels of ionization.
Understanding the nature of triprotic acids is crucial when studying acid-base reactions, as it influences how these substances behave in chemical environments, such as pH regulation and buffer systems.

Dissociation Reactions

Dissociation reactions involve the breaking apart of an acid in water to produce ions. For a triprotic acid like citric acid, this occurs in three stages, each involving the release of a proton, and formation of a conjugate base.
At the first stage, one proton detaches:

• Citric acid (H₃C₆H₅O₇) dissociates to form its first conjugate base (H₂C₆H₅O₇⁻) and a hydronium ion (H₃O⁺).

The second stage sees another proton removed from this first conjugate base, yielding the second conjugate base:

• First conjugate base (H₂C₆H₅O₇⁻) turns into the second conjugate base (HC₆H₅O₇²⁻).

Finally, in the third stage, the last proton is released:

• The second conjugate base (HC₆H₅O₇²⁻) forms the third and final conjugate base (C₆H₅O₇³⁻).

Each step is associated with the formation of a hydronium ion, affecting the pH of the solution.

Conjugate Base

When an acid loses a proton during a dissociation reaction, what remains is the conjugate base. With citric acid, each of the dissociation stages forms a different conjugate base. This is fundamental to understanding how triprotic acids dissociate.
In the first reaction step, citric acid ( H₃C₆H₅O₇) loses a proton, turning into its first conjugate base (H₂C₆H₅O₇⁻).
The second step involves this first conjugate base releasing another proton to form the second conjugate base (HC₆H₅O₇²⁻).
Finally, the third reaction sees the second conjugate base give up one more proton, resulting in the third and final conjugate base (C₆H₅O₇³⁻).

• The conjugate bases become more negatively charged at each step, influencing their reactivity.
• This stepwise formation affects equilibrium dynamics, an important aspect in chemical buffering.

Each base formed is crucial for calculating acidic outcomes, such as pH levels.

Acid Dissociation Constant (Ka)

The acid dissociation constant, abbreviated as Ka, is a measure of the strength of an acid in solution. It provides information on the equilibrium between the dissociated ions and the un-dissociated molecules present in solution.
Citric acid, being triprotic, has three different Ka values, one for each dissociation step ( K_{a1}, K_{a2}, K_{a3}). These constants reflect the degree of ionization for each stage:

• Ka1: Represents the dissociation of the first proton and is usually the highest among the three, indicating greater readiness to release the first proton.
• Ka2: Deals with the second proton's dissociation, typically showing a smaller value due to decreased tendency to lose another proton from the first conjugate base.
• Ka3: Reflects the dissociation of the third proton, displaying the smallest value as it is even harder to remove the last proton from the second conjugate base.

These constants are crucial because they help predict how an acid will behave in the solution, especially its reactivity and the resulting pH when in contact with other substances.