Sulfurous acid ( H2SO3) is a diprotic acid with a1=1.39×10−2 and a2=6.73×10−8.
Determine the pH of a 0.189M sulfurous acid ( H2SO3 ) solution.

Answers

Answer 1

A sulfuric acid solution with a concentration of 0.189M has a pH of roughly 0.778.

Sulfuric acid: Is it a diprotic acid?

A diprotic acid is sulfuric acid, Sulfurous acid. The findings of an acid-base titration can be used to calculate the values of Ka1 and Ka2. To completely neutralise diprotic acids, two hydroxide ions are needed for each molecule. Sulfuric acid, a stronger acid than Hydrogen sulfide, with a pH of 1.5 in a solution of 0.100 mol dm³.

The chemical equation for sulfuric acid dissociation is as follows:

Sulfuric acid + Water ⇌ Hydronium ion + hydrosulfite anion

Write the expression for the acid dissociation constant (Ka1) for the first dissociation:

Ka1 = [Hydronium ion[hydrosulfite anion]/[Sulfuric acid]

Substitute the given value of Ka1 (1.39×10−2) and the initial concentration of Sulfuric acid (0.189 M) into the expression for Ka1 and solve for [hydronium ion]:

1.39×10−2 = [hydronium ion][Hydrogen sulfite]/0.189

[hydronium ion] = 0.167 M

pH = -log[hydronium ion]

pH = -log(0.167)

pH = 0.778

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