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The pKₐ of sulfurous acid in the following reaction is 1.82 - HSC - SSCE Chemistry - Question 36 - 2021 - Paper 1

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The pKₐ of sulfurous acid in the following reaction is 1.82. H₂SO₃(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HSO₃⁻(aq) The pKₐ of hydrogen sulfite in the following reaction is 7.1... show full transcript

Worked Solution & Example Answer:The pKₐ of sulfurous acid in the following reaction is 1.82 - HSC - SSCE Chemistry - Question 36 - 2021 - Paper 1

Step 1

pKₐ of sulfurous acid to Kₐ

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Answer

The first step is to convert the pKₐ of sulfurous acid to its equilibrium constant Kₐ:

Ka(1)=10−pKa=10−1.82=1.51×10−2K_{a(1)} = 10^{-pKₐ} = 10^{-1.82} = 1.51 \times 10^{-2}

Step 2

pKₐ of hydrogen sulfite to Kₐ

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Answer

Next, convert the pKₐ of hydrogen sulfite:

Ka(2)=10−pKa=10−7.17=6.76×10−8K_{a(2)} = 10^{-pKₐ} = 10^{-7.17} = 6.76 \times 10^{-8}

Step 3

Combine Kₐ for the overall reaction

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Answer

For the given reaction:

H₂SO₃(aq) + 2H₂O(l) ⇌ 2H₃O⁺(aq) + SO₃²⁻(aq)

We can express the equilibrium constant K_eq as:

Keq=Ka(1)×Ka(2)K_{eq} = K_{a(1)} \times K_{a(2)}

Substituting the values:

Keq=(1.51×10−2)×(6.76×10−8)K_{eq} = (1.51 \times 10^{-2}) \times (6.76 \times 10^{-8})

Calculating this gives us:

Keq=1.01×10−9K_{eq} = 1.01 \times 10^{-9}

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