You are given a solution of lead nitrate. In order to obtain a yellow precipitate you should mix it with a solution of:

Potassium chloride

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Potassium nitride

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Potassium sulphide

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Potassium iodide

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Solution

The correct option is D Potassium iodide
Potassium iodide reacts with lead nitrate to give a yellow ppt. of lead iodide.
The reaction takes place as follows,
$K I (a q) Potassium iodide + P b (N O_{3})_{2} Lead nitrate \to P b I_{2} Lead iodide(s) + 2 K N O_{3} (a q)$

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Similar questions

Q. You are given the solution of lead nitrate. In order to obtain a yellow precipitate you should mix with it a solution of:
(a) potassium chloride
(b) potassium nitride
(c) potassium sulphide
(d) potassium iodide

Q. Assertion :"A golden yellow precipitate of lead iodide is formed when colourless lead nitrate solution is mixed with colourless solution of potassium iodide." So, it is a chemical change Reason: Formation of new products in the reaction with the precipitate indicates a chemical change

$P b (N O_{3})_{2} (a q) + 2 K C l (a q) \to 2 K N O_{3} (a q) + P b C l_{2} (s)$

When lead nitrate solution is mixed with potassium chloride solution, a precipitate of lead chloride and potassium nitrate solution are formed. This type of reaction is called _________________.

Q. You are given a colourless solution and was told to add a little

N a C l

solution to it. A cloudy white precipitate was made. What conclusion should you draw about the nature of the colourless solution? The ppt dissolves in hot water, then the ppt is :

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

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You are given a solution of lead nitrate. In order to obtain a yellow precipitate you should mix it with a solution of:

The correct option is D Potassium iodidePotassium iodide reacts with lead nitrate to give a yellow ppt. of lead iodide. The reaction takes place as follows, KI(aq)Potassium iodide+Pb(NO3)2Lead nitrate→PbI2Lead iodide(s)+2KNO3(aq)

The correct option is D Potassium iodide
Potassium iodide reacts with lead nitrate to give a yellow ppt. of lead iodide.
The reaction takes place as follows,
$K I (a q) Potassium iodide + P b (N O_{3})_{2} Lead nitrate \to P b I_{2} Lead iodide(s) + 2 K N O_{3} (a q)$