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. $K_a=\frac{[HAsO_4^{2-}]\times [H_3O^+]}{[H_2AsO_4^{-}]}$

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:

$K_a=\frac{[HAsO_4^{2-}]\times [H_3O^+]}{[H_2AsO_4^-]}$