Answer:
C.) [tex]K_a = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]}[/tex]
Explanation:
The general structure for a Ka expression is:
[tex]K_a = \frac{[H^+][A^-]}{[HA]}[/tex]
In this expression,
-----> Ka = equilibrium constant
-----> [A⁻] = base
-----> [HA] = acid
The products are in the numerator and the reactants are in the denominator. In this case, H₃PO₄ serves as an acid and H₂PO₄⁻ serves as a base.
As such, the equilibrium expression is:
[tex]K_a = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]}[/tex]
