Answer
Equation for the reaction:
$$H_3PO_4(aq) + H_2O(l) \leftrightharpoons H_2{PO_4}^-(aq) + H_3O^+(aq)$$
Acid dissociation constant expression:
$$K_a = \frac{[H_3O^+][H_2P{O_4}^-]}{[H_3PO_4]} = 7.5 \times 10^{-3}$$
Work Step by Step
1. Write the reactants and products:
Phosphoric acid : $H_3PO_4$
Water (dissociation): $H_2O$
Dihydrogen phosphate: $H_2{PO_4}^-$
Hydronium ion: $H_3O^+$
2. Write the equation:
$$H_3PO_4(aq) + H_2O(l) \leftrightharpoons H_2{PO_4}^-(aq) + H_3O^+(aq)$$
3. The $K_a$ expression follows this pattern:
$$K_a = \frac{[H_3O^+][Conj. \space Base]}{[Acid]}$$
Substituting the compounds:
$$K_a = \frac{[H_3O^+][H_2P{O_4}^-]}{[H_3PO_4]} = 7.5 \times 10^{-3}$$
