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(a) What happens when : (i) sulphur dioxide is passed through an aqueous solution of Fe (III) salt ? (ii) Sulphur dioxide is passed through and aqueous solution of sodium carbonate ? (b) On passing H_(2)S through an aqueous solution of SO_(2), as a yellow turbidity is formed. Why ? |
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Answer» Solution :(a) (i) `SO_(2)` being a reducing agent, reduces Fe (III) to Fe (II) salts. `2overset(3+)(Fe)(aq)+SO_(2)(g) + 2H_(2)O (l) rarr 2 overset(2+)(Fe)(aq) + SO_(4)^(2-) (aq) + 4 overset(+)(H)(aq)` (ii) Being acidic, `SO_(2)` decomposes `Na_(2)CO_(3)` FORMING sodium SULPHITE with evolution of `CO_(2)` gas. `Na_(2)CO_(3)(aq) + SO_(2)(g) rarr Na_(2)SO_(3) (aq) + CO_(2)(g)` (B) `H_(2)S` reduces `SO_(2)` to sulphur which appears as yellow TURBIDITY. `2H_(2)S + SO_(2) rarr underset(("Yellow turbidity"))(3S) + 2H_(2)O` |
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