Question

The given solution of lead nitrate, in order to obtain a yellow precipitate, must be mixed with:

• A. Potassium chloride
• B. Potassium nitrite
• C. Potassium iodide
• D. Potassium sulfate

Answer 1

The correct option is C

Potassium iodide

The explanation for the correct option:

(C) Potassium Iodide

${\mathrm{PbI}}_2$(yellow ppt.) is formed when lead nitrate and potassium iodide are combined.

$\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{aq}\right)+2\mathrm{KI}\left(\mathrm{aq}\right)\to {\mathrm{PbI}}_2\left(\mathrm{s}\right)+2{\mathrm{KNO}}_3\left(\mathrm{aq}\right)$

The explanation for the incorrect option(s):

Option (A)

1. Lead chloride formed is a white crystalline powder.
2. $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{aq}\right)+\mathrm{KCl}\left(\mathrm{aq}\right)\to {\mathrm{PbCl}}_2\left(\mathrm{s}\right)+{\mathrm{KNO}}_3\left(\mathrm{aq}\right)$

Option (B)

1. Reaction does not take place.
2. $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{aq}\right)+{\mathrm{KNO}}_3\left(\mathrm{aq}\right)\to \mathrm{No}\mathrm{reaction}$

Option (D)

1. Lead sulfate formed is white powder.
2. $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{aq}\right)+{\mathrm{K}}_2{\mathrm{SO}}_4\left(\mathrm{s}\right)\to 2{\mathrm{KNO}}_3\left(\mathrm{aq}\right)+{\mathrm{PbSO}}_4\left(\mathrm{s}\right)$

Hence, the correct option is (C).