Sulfurous acid is a polyprotic acid. write balanced chemical equations for the sequence of reactions that sulfurous acid can undergo when it's dissolved in water.

The balanced chemical equations for the acid dissociation of sulfurous acid are:

Sulfurous acid is a polyprotic acid.

A polyprotic acid is an acid that can release more than one proton in an aqueous solution.

Sulfurous acid has 2 protons so it releases them in 2 steps.

H₂SO₃(aq) ⇄ HSO₃⁻(aq) + H⁺(aq)

HSO₃⁻(aq) ⇄ SO₃²⁺(aq) + H⁺(aq)

The balanced chemical equations for the acid dissociation of sulfurous acid are:

Learn more about polyprotic acids here: link

The reactions are:

H₂SO₃  +  H₂ O  ⇄  HSO₃⁻  +  H₃O⁺

HSO₃⁻  +  H₂O  ⇄  SO₃⁻²  +  H₃O⁺

Explanation:

This is the sulfurous acid → H₂SO₃.

It is a weak diprotic acid and it is diprotic because it can release 2 protons to increase H₃O⁺ from water. As every weak acid, the reactions are in equilibrium, that's why it has two Ka. As it can release protons, it is an acid from Bronsted Lowry and an acid from Arrhenius.

H₂SO₃  +  H₂ O  ⇄  HSO₃⁻  +  H₃O⁺   Ka₁

 HSO₃⁻  → this is called acid sulfite

HSO₃⁻  +  H₂O  ⇄  SO₃⁻²  +  H₃O⁺     Ka₂

2H₂O  ⇄   H₃O⁺  +  OH⁻    Kw

(Don't forget to put water equation, always)

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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