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In aqueous solution, sulphuric acid ionizes in two steps: H 2 SO 4 is very strong in water largely because of its first ionisation to H 3 O + and HSO 4 – . H 2 SO 4 (aq) +H 2 O(l) → H 3 O + (aq) + HSO 4 - (Ka 1 HSO 4 - + H 2 O(l) → H 3 O + +SO 4 -2 (Ka 2 The ionisation of HSO 4 – to H 3 O – and SO 4 2– is very small. That is why Ka 2 < < Ka 1 ,.

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The p-Block Elements

Short Answer Type

391. Why does O₃ act as a powerful oxidising agent?

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392. How is O₃ estimated quantitatively?

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393. What happens when sulphur dioxide is passed into aqueous solution of Fe (III) salt?

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394. Comment on the nature of two S-O bonds formed in SO₂ molecule. Are the two S-O bonds in this molecule equal?

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395. How is the presence of SO₂ detected?

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396. Mention three areas in which H₂SO₄ plays an important role.

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397. Write the conditions to maximise the yield of H₂SO₄ by contact process.

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398. Why is Ka₂ < < Ka, for H₂SO₄ in water?

In aqueous solution, sulphuric acid ionizes in two steps:
H₂ SO₄ is very strong in water largely because of its first ionisation to H₃O⁺ and HSO₄^–.

H₂SO₄(aq) +H₂O(l)→ H₃O⁺ (aq) + HSO₄^- (Ka₁
HSO₄^- + H₂O(l) → H₃O⁺ +SO₄^-2 (Ka₂

The ionisation of HSO₄^– to H₃O^– and SO₄^2– is very small. That is why Ka₂ < < Ka₁,.

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399. Considering the parameters such as bond dissociation enthalpy, electron gain enthalphy and hydration enthalpy compare the oxidising power of F₂ and Cl₂.

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