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

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The $pK_a$ of sulfurous acid in the following reaction is 1.82. $$ H_2SO_3(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + HSO_3^-(aq) $$ The $pK_a$ of hydrogen sulf... show full transcript

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

Step 1

Calculate $K_{a(1)}$ for sulfurous acid:

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Answer

Ka(1)=10−pKa=10−1.82=0.01513651248≈1.51×10−2K_{a(1)} = 10^{-pK_a} = 10^{-1.82} = 0.01513651248 \approx 1.51 \times 10^{-2}

Step 2

Calculate $K_{a(2)}$ for hydrogen sulfite:

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Answer

Ka(2)=10−pKa=10−7.17=0.0000000676≈6.76×10−8K_{a(2)} = 10^{-pK_a} = 10^{-7.17} = 0.0000000676 \approx 6.76 \times 10^{-8}

Step 3

Determine $K_{eq}$ for the reaction:

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For the reaction:

H2SO3(aq)+2H2O(l)⇌2H3O+(aq)+SO32−(aq)H_2SO_3(aq) + 2H_2O(l) \rightleftharpoons 2H_3O^+(aq) + SO_3^{2-}(aq) We have:

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)≈1.0×10−9K_{eq} = (1.51 \times 10^{-2}) \times (6.76 \times 10^{-8}) \approx 1.0 \times 10^{-9}

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