Question

Write the acidic equilibrium equation for HPO42

Answer 1

The equation for HPO42- as a weak acid is HPO42-(aq) + H2O(l) → H3O+(aq) + PO43-(aq), and its acid dissociation constant (Ka) is given by Ka = [H3O+][PO43-] / [HPO42-], with a known value of 4.2 × 10-13.

Step-by-step explanation:

The acidic equilibrium equation for HPO42- acting as a weak acid in water is written as follows:

HPO42-(aq) + H2O(l) → H3O+(aq) + PO43-(aq)

The acid dissociation constant (Ka) expression for this reaction, where HPO42- donates a proton (H+) to form the conjugate base PO43- and hydronium ion H3O+, is represented as:

Ka = [H3O+][PO43-] / [HPO42-]

The value of Ka for this reaction, known as the second ionization constant of phosphoric acid (H3PO4), is typically much smaller than the first ionization constant, indicating a weaker degree of ionization in this step. The given value for Ka is 4.2 × 10-13, which helps in calculating equilibrium concentrations in a solution of H3PO4.

Answer 2

Step-by-step explanation:

`HPO_4^(2-)` is a conjugate base of
`H_2PO_4^(-)` but it has one acidic hydrogen which dissociates in the solution and exists in the equilibrium as shown below:

`HPO_4^(2-)+H_2O\rightleftharpoons H_3O^++PO_4^(3-)`

The expression for the dissociation constant,
`K_a` for
`HPO_4^(2-)` is shown below as:

`K_a=\frac {[H_3O^+][PO_4^(3-)]}{[HPO_4^(2-)]}`