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Question

In the double displacement reaction between aqueous potassium iodide and aqueous lead nitrate, a yellow precipitate of lead iodide is formed. While performing the activity if lead nitrate is not available, which of the following can be used in place of lead nitrate?

A
Lead sulphate (insoluble)
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B
Lead acetate
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C
Ammonium nitrate
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D
Potassium sulphate
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Solution

The correct option is A Lead acetate
2KI(aq)+Pb(NO3)2(aq)PbI2(s)+2KNO3
In the above double displacement reaction, potassium Iodide(KI) and lead nitrate (Pb(NO3)2) dissociate in their aqueous states to form ions. The lead (Pb2+) ions combine with the iodide (I) ions to form precipitates of lead iodide (PbI2). If Lead sulphate is used in place of lead nitrate, no precipitates of lead iodide will be formed, because lead sulphate being insoluble in water does not give Pb2+ ions which can combine with I to form PbI2. On the other hand, lead acetate dissociates in aqueous state to give Pb2+ ions and CH3COO ions . Therefore potassium iodide combines with lead acetate to form precipitates of lead iodide and potassium acetate.
2KI(aq)+Pb(CH3COO)2(aq)PbI2(s)+2CH3COOK

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