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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
$$2KI(aq)  +  Pb(NO_{3})_{2}(aq)  \rightarrow   PbI_{2}(s)  +  2KNO_{3}$$
In the above double displacement reaction, potassium Iodide(KI) and lead nitrate ($$Pb(NO_{3})_{2}$$) dissociate in their aqueous states to form ions. The lead ($$Pb^{2+}$$) ions combine with the iodide ($$I^{-}$$) ions to form precipitates of lead iodide ($$PbI_{2}$$). 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 $$Pb^{2+}$$ ions which can combine with $$I^{-}$$ to form $$ PbI_{2}$$. On the other hand, lead acetate dissociates in aqueous state to give $$Pb^{2+}$$ ions and $$CH_{3}COO^{-}$$ ions . Therefore potassium iodide combines with lead acetate to form precipitates of lead iodide and potassium acetate.
$$2KI(aq)  +  Pb(CH_{3}COO)_{2}(aq)  \rightarrow   PbI_{2}(s)  +  2CH_{3}COOK$$

Chemistry

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