Question

Write the equilibrium constant expression for this reaction: 3h (aq) po−34 (aq) → h3po4 (aq)

Answer 1

The equilibrium constant expression for the given ionization reaction of phosphoric acid involves the ratio of the concentration of H3PO4 to the cube of the concentration of H+ and the concentration of PO43-. To calculate the actual Keq, concentrations of all species at equilibrium would be needed.

Step-by-step explanation:

The equilibrium constant expression for the given reaction, 3H+ (aq) + PO43- (aq) → H3PO4 (aq), is defined by the concentrations of the products over the reactants, each raised to the power of their stoichiometric coefficients in the balanced equation. This reaction represents the third ionization of phosphoric acid, H3PO4.

The equilibrium constant expression, Keq, is given as:

Keq = [H3PO4]/([H+]3 · [PO43-])

Given the composition of an equilibrium mixture as follows: [H+] = 1.3 × 10-6 M, [H2PO4−] = 0.042 M, and [HPO42−] = 0.341 M, we would need the concentration of H3PO4 to calculate the equilibrium constant for this particular system, which is not provided in the scenario. Finally, note that the large differences in successive ionization constants for polyprotic acids (like phosphoric acid) simplifies equilibrium calculations since we usually consider only the predominant species at each ionization stage.

Answer 2

Answer:
`K=([H_3PO_4])/([H^+]^3[PO_4^(3-)])`

Step-by-step explanation:

Equilibrium constant is the ratio of product of concentration of products to the product of concentration of reactants each term raise to their stochiometric coefficients.

Thus for the reaction:

`3H^+(aq)+PO_4^(3-)(aq)\rightarrow H_3PO_4(aq)`

`K=([H_3PO_4])/([H^+]^3[PO_4^(3-)])`