For which of the following equilibria does Kc correspond to the acid-dissociation constant, Ka, of H2PO4-?

A) H2PO4-(aq) + H3O+(aq)--> H3PO4(aq) + H2O(l)
B) H2PO4-(aq) + H2O(l)--> H3PO4(aq) + OH-(aq)
C)H2PO4-(aq) + H2O(l)--> H3O+(aq) + HPO42(aq)
D) H3PO4(aq) + H2O(l)--> H3O+(aq) + H2PO4-(aq)
E) HPO42-(aq) + H2O(l)--> H2PO4-(aq) + OH-(aq)

Respuesta :

Answer:

The correct answer is C)H2PO4-(aq) + H2O(l)--> H3O+(aq) + HPO42-(aq)

Explanation:

The acid dissociation equilibrium involves the loss of a proton of the acid to give the conjugated acid. In this case, the acid is H₂PO₄⁻ and it losses a proton (H⁺) to give the conjugated acid HPO₄²⁻ (without a proton and with 1 more negative charge). In the aqueous equilibrium, the proton is taken by H₂O molecule to give the hydronium ion H₃O⁺.

H₂PO₄⁻(aq) + H₂O(l)--> H₃O⁺(aq) + HPO₄²⁻(aq)

Answer:

C) H₂PO₄⁻(aq) + H₂O(l) → H₃O⁺(aq) + HPO₄²⁻(aq)

Explanation:

For which of the following equilibria does Kc correspond to the acid-dissociation constant, Ka, of H₂PO₄⁻?

A) H₂PO₄⁻(aq) + H₃O⁺(aq) → H₃PO₄(aq) + H₂O(l)

NO. This is the inverse of the acid dissociation of H₃PO₄.

B) H₂PO₄⁻(aq) + H₂O(l) → H₃PO₄(aq) + OH⁻(aq)

NO. This is the basic dissociation of H₂PO₄⁻.

C) H₂PO₄⁻(aq) + H₂O(l) → H₃O⁺(aq) + HPO₄²⁻(aq)

YES. This is the acid dissociation of H₂PO₄⁻. The acid-dissociation constant is:

[tex]Ka=\frac{[H_{3}O^{+}].[HPO_{4}^{2-} ]}{[H_{2}PO_{4}^{-} ]}[/tex]

D) H₃PO₄(aq) + H₂O(l) → H₃O⁺(aq) + H₂PO₄⁻(aq)

No. This is the acid dissociation of H₃PO₄.

E) HPO₄²⁻(aq) + H₂O(l) → H₂PO₄⁻(aq) + OH⁻(aq)

NO. This is the basic dissociation of HPO₄²⁻.