Question

Write the balanced equation and Ka expression for the Bronsted-Lowery acid H2AsO4 -- in water?

Answer 1

Answer: a.
`H_2AsO_4^(-)(aq)+H_2O(l)\rightarrow HAsO_4^(2-)(aq)+H_3O^+(aq)`

b.
`K_a=([HAsO_4^(2-)]* [H_3O^+])/([H_2AsO_4^(-)])`

Explanation:-

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

The balanced chemical equation is:

`H_2AsO_4^(-)(aq)+H_2O(l)\rightarrow HAsO_4^(2-)(aq)+H_3O^+(aq)`

Here,
`H_2AsO_4^(-)` is loosing a proton, thus it is considered as an acid and after losing a proton, it forms
`HAsO_4^(2-)` which is a conjugate base.

And,
`H_2O` is gaining a proton, thus it is considered as a base and after gaining a proton, it forms
`H_3O^+` which is a conjugate acid.

The dissociation constant is given by:

`K_a=([HAsO_4^(2-)]* [H_3O^+])/([H_2AsO_4^-])`