Question

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 sulfide
D. potassium iodide

Answer 1

We will determine the product of the reaction and check their solubility. The compound which will be insoluble in water will form a precipitate. In the reaction between lead nitrate and potassium halides, lead halides will be formed. We know that halides except $PbC{{l}_{2}},H{{g}_{2}}C{{l}_{2}},AgCl$ are insoluble in water. So, they will form precipitate. Hence, we will see the reactions, the precipitate formed and then we can solve the problem, if we remember the color of certain ions in precipitate.

Complete step-by-step answer: Let us see all reactions by reacting lead nitrate with all the potassium salts one by one. We will analyze the ion present in the precipitate and then will deduce the color of precipitate.
A. Reaction of lead nitrate with potassium chloride.
The balanced chemical equation for the reaction is as follows;
$Pb{{\left( N{{O}_{3}} \right)}_{2}}\left( aq \right)+2KCl\left( aq \right)\to PbC{{l}_{2}}\left( s \right)\downarrow +2KN{{O}_{3}}\left( aq \right)$
We can see that lead chloride (II) is formed as a product. And we know that lead chloride (II) is the only lead halide which will be insoluble in water, hence it will form precipitate, which is of white color.
B. Reaction of lead nitrate with potassium nitride.
This reaction in general does not take place.
C. Reaction of lead nitrate and potassium sulfide
The balanced chemical equation for the following is represented as
${{K}_{2}}S\left( aq \right)+Pb{{\left( N{{O}_{3}} \right)}_{2}}\left( aq \right)\to 2KN{{O}_{3}}\left( aq \right)+PbS\left( s \right)$
Here, we observe that lead sulfide is formed as a product. It is a black ppt.
D. Reaction of lead nitrate with potassium iodide.
The balanced chemical equation for the following reaction can be represented as
$2KI\left( aq \right)+Pb{{\left( N{{O}_{3}} \right)}_{2}}\left( aq \right)\to Pb{{I}_{2}}\left( s \right)+2KN{{O}_{3}}$
Here, lead iodide (II) is formed as a product. We should remember that lead iodide is insoluble in water and forms ppt. The color of ppt of lead iodide (II) is yellow.
So, we can say that when lead nitrate reacts with potassium iodide it forms a yellow ppt of lead chloride (II)

Hence, the correct option is option D.

Note: All the above reactions are double displacement reactions, in which a higher reactive element displaces a lower reactive element from its salt solution. One should learn the names of some common precipitate colors. This proves to be very helpful.