Final answer:
The Kb for HPO₄²⁻ is derived from its reaction with water and the given Ka value. By applying the formula Kw = Ka × Kb, where Kw is the ion product constant of water, the Kb can be calculated.
Step-by-step explanation:
Expression for Kb of HPO₄²⁻
The expression for the Kb for the anion HPO₄²⁻, which is amphiprotic, can be derived from its reaction with water where it acts as a base. The equilibrium equation for this reaction is:
HPO₄²⁻(aq) + H₂O(l) ⇌ H₂PO₄⁻(aq) + OH⁻(aq)
The equilibrium constant for this reaction is known as the base ionization constant, Kb, and is expressed as:
Kb = [H₂PO₄⁻][OH⁻] / [HPO₄²⁻]
For the given ion HPO₄²⁻, since it can also act as an acid, we have the Ka value. However, since HPO₄²⁻ is primarily acting as a base here, we are interested in the Kb. The Ka value for HPO₄²⁻ given is 2.2 × 10⁻¹³. To find the Kb value, we need to use the relationship between Ka and Kb which is Kw = Ka × Kb, where Kw is the ion product constant of water at 25°C and equals to 1.0 × 10⁻.
To solve for Kb of HPO₄²⁻:
Kb = Kw / Ka = 1.0 × 10⁻ / 2.2 × 10⁻ ≈ 4.545 × 10⁻