Answer :
Answer:
Kb = 1.587x10⁻⁷
Explanation:
The equilibriums of phosphoric acid in water are:
H₃PO₄ ⇄ H₂PO₄⁻ + H⁺ Ka = 7.2x10⁻³
H₂PO₄⁻ ⇄ HPO₄²⁻ + H⁺ Ka = 6.3x10⁻⁸
HPO₄²⁻ ⇄ PO₄³⁻ + H⁺ Ka = 4.2x10⁻¹³
The Kb of HPO₄²⁻ is:
HPO₄²⁻ + H₂O ⇄ H₂PO₄⁻ + OH⁻
This equation is the result of the water equilibrium - Second Ka of phosphoric acid.
And Kb = Kw = Ka
Kb = 1x10⁻¹⁴ / 6.3x10⁻⁸
Kb = 1.587x10⁻⁷
Given that H₃PO₄ is a triprotic acid for which Ka1 = 7.2 × 10⁻³, Ka2 = 6.3 × 10⁻⁸ and Ka3 = 4.2 × 10⁻¹³, the Kb for HPO₄²⁻ is 1.6 × 10⁻⁷.
Let's consider the 3 steps for the dissociation of phosphoric acid.
Step 1: H₃PO₄ + H₂O ⇒ H₂PO₄⁻ + H₃O⁺ Ka1
Step 2: H₂PO₄⁻ + H₂O ⇒ HPO₄²⁻ + H₃O⁺ Ka2
Step 3: HPO₄²⁻ + H₂O ⇒ PO₄³⁻ + H₃O⁺ Ka3
As we can see in the second step, HPO₄²⁻ is the conjugate base of H₂PO₄⁻. We can calculate its basic dissociation constant (Kb) using the following expression.
[tex]Kw = Ka2 \times Kb\\\\Kb = \frac{Kw}{Ka2} = \frac{1.0 \times 10^{-14} }{6.3 \times 10^{-8} } = 1.6 \times 10^{-7}[/tex]
Given that H₃PO₄ is a triprotic acid for which Ka1 = 7.2 × 10⁻³, Ka2 = 6.3 × 10⁻⁸ and Ka3 = 4.2 × 10⁻¹³, the Kb for HPO₄²⁻ is 1.6 × 10⁻⁷.
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