Question

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

Answer 1

• When Potassium iodide is added with Lead nitrate, the Iodide ion displaces nitrate from lead nitrate and nitrate displaces iodine from potassium iodide. $\underset{\mathrm{Lead}\mathrm{nitrate}}{\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{aq}\right)}+\underset{\mathrm{Potassium}\mathrm{iodide}}{2\mathrm{KI}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Lead}\mathrm{iodide}(\mathrm{Yellow}\mathrm{precipitate})}{{\mathrm{PbI}}_2\left(\mathrm{s}\right)}+\underset{\mathrm{Potassium}\mathrm{nitrate}}{2{\mathrm{KNO}}_3\left(\mathrm{aq}\right)}$
• 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 ${\mathrm{PbI}}_2$ (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.