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Question

Reema took 5ml of Lead nitrate solution in a beaker and added approximately 4ml Potassium iodide solution to it. What would she observe?


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Solution

  • When Potassium iodide is added with Lead nitrate, the Iodide ion displaces nitrate from lead nitrate and nitrate displaces iodine from potassium iodide. Pb(NO3)2(aq)Leadnitrate+2KI(aq)PotassiumiodidePbI2(s)Leadiodide(Yellowprecipitate)+2KNO3(aq)Potassiumnitrate
  • As a result, two new products are formed Potassium nitrate and Lead iodide. Hence, this is a double displacement reaction.
  • A yellow-colored precipitate of Lead iodide is generated when Potassium iodide solution is introduced to the Lead nitrate solution.
  • Potassium nitrate dissolves in water. Lead iodide, on the other hand, is only moderately soluble in water. The majority of the Lead iodide in the solution precipitates out as a yellow solid.
  • The reaction between Potassium iodide and Lead nitrate is an example of a precipitation reaction since it produces the precipitate PbI2 (lead iodide).

Therefore, Reema observed that two new products are formed Potassium nitrate and Lead iodide due to a double displacement reaction, and Potassium Iodide is precipitated.


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