Question

Write chemical
Q-1 Name two solutions which on mixing form a yellow precipitate,
equation involved in it.​

Answer 1

Answer:

2KI(aq)+Pb(NO

3

)

2

(aq)→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)→PbI

2

(s)+2CH

3

COOK

Explanation:

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Answer 2

The chemical reaction between Potassium iodide and lead nitrate is characterized by the formation of a yellow precipitate of lead iodide.

• A precipitate is a solid product which separates from the solution during a chemical reaction.

>> This is an example of a double displacement reaction.

• Those reactions in which two compounds react by the exchange of ions to form two new compounds are called double displacement reactions.

>> The chemical equation for this reaction is,

$\tt Pb(NO_{3}) _{2} (aq) \:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:2KI(aq)\:\:\:\: \:\:\:\: \:\:\:\:\longrightarrow \:\:\:\:\:\:\:\:PbI_{2}(s)\:\:\:\: \:\:\:\:+\:\:\:\: \:\:\:\:2KNO_{3}(aq) \\\tt Lead\: nitrate \:\:\:\:\:\:\:\:\:\:\:\:Potassium \:iodide\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:Lead\: iodide\:\:\:\:\:\:\:\:\:\:\:\:Potassium \: nitrate$