Question

addition of $KI$ to $Pb$ salt in water gives ---- ppt:
A. Yellow
B. Black
C. White
D. Red

Answer 1

Hint:In qualitative analysis of inorganic ionic compounds, we separately detect anions and cations. First the anion and then the cationic part. $Pb$ comes in the first as well as second group in qualitative analysis of cations.

Complete answer:
Qualitative analysis is a method of analytical chemistry that deals with the determination of elemental composition of inorganic salts. It is mainly concerned with the detection of ions in aqueous solution of the salt.
The common method for testing is that first we make a solution and then will test this solution with reagents for particular ions that may be present in the given solution. Reaction of various reagents with ions shows color change, or a precipitate will form or any other visible change.
In cation analysis, Some tests are preliminary which we do before going stepwise in the groups, like flame test, cobalt nitrate test, physical appearance etc.
In the zero group ammonium ion comes only.
In the first group, it includes lead ion, silver ion, mercury ion. The group reagent for his group is $HCl $. The addition of $HCl$ will precipitate lead ion as $Pb{I_2}$ .
When $PbC{l_2}$ solution in hot water reacts with potassium iodide, $KI$ results in yellow precipitate of lead iodide. Reaction is as follows:
$PbC{l_2} + 2KI \to Pb{I_2} \downarrow + 2KCl$
(hot solution) $\:\:\:$ (yellow ppt)
This test is known as potassium iodide test. Here double displace reaction occurs, in which lead and potassium ions switch partners to form soluble iodide $KCl$ , and $Pb{I_2}$ which is yellow solid that precipitates out of the solution because it is insoluble in water.

Hence the correct option is A. Yellow

Note:
Another test for the confirmation of lead is potassium chromate test, in which hot solution of lead chloride reacts with potassium chromate which leads to the formation of yellow precipitate of lead chromate. Lead ion also comes in group II.