
The reagent that distinguishes between silver and lead salt is :
(A) ${H_2}S$ gas
(B) Dil. $HCl$ solution added after this dissolved in hot water.
(C) $N{H_4}Cl\left( {solid} \right) + N{H_4}OH\left( {solution} \right)$
(D) $N{H_4}Cl\left( {solid} \right) + \left( {N{H_4}} \right){ _2}C{O_3}\left( {solution} \right)$
Answer
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Hint: Find a suitable reagent that precipitates with one of the compounds in order to discriminate between two compounds; the reagent should not precipitate with the other component. If the reagent precipitates one of the compounds with both compounds present, only one of the precipitates should dissolve in the excess reagent.
Complete Step by Step Solution:
Since we are using the concept of precipitation, we must know its exact meaning as well. Precipitation is the process of changing a dissolved substance from a saturated solution to a non-soluble solid in an aqueous solution. The solid substance so formed is known as precipitate.
Also, whether a substance will precipitate in a given solvent or not will depend on the solubility laws along with the exceptions involved.
Silver belongs to group 11 of the periodic table and lead belongs to group 14.
In the given question we need to tell which of the given reagents can distinguish silver and lead salts.
Silver and lead will precip itate out as silver and lead chlorides respectively when diluted $HCl$ solution is added. Now, if we combine them with hot water, the resulting $AgCl$ will not dissolve in water whereas the resulting $PbC{l_2}$ will. This is due to the fact that $PbC{l_2}$ is soluble in hot water but insoluble in cold water, unlike $AgCl$ , which is insoluble in both conditions. $Pb\,\& \,Ag$ both produce white precipitate from their chlorides. However, $AgCl$ does not dissolve in hot water but $PbC{l_2}$ does.
Thus, dil.HCl solution added after this dissolved in hot water serves as the reagent that distinguishes between silver and lead salt.
Consequently, they can be separated in this manner.
Hence, option B. is the right answer.
Note: When the ionic product is greater than the solubility product, a precipitate is produced. The production of a specific precipitate typically indicates the presence of a specific ion in qualitative analyses of inorganic ions. For instance, you can say that the unknown contains chloride ions if you get silver chloride precipitate after treating it with silver nitrate.
Complete Step by Step Solution:
Since we are using the concept of precipitation, we must know its exact meaning as well. Precipitation is the process of changing a dissolved substance from a saturated solution to a non-soluble solid in an aqueous solution. The solid substance so formed is known as precipitate.
Also, whether a substance will precipitate in a given solvent or not will depend on the solubility laws along with the exceptions involved.
Silver belongs to group 11 of the periodic table and lead belongs to group 14.
In the given question we need to tell which of the given reagents can distinguish silver and lead salts.
Silver and lead will precip itate out as silver and lead chlorides respectively when diluted $HCl$ solution is added. Now, if we combine them with hot water, the resulting $AgCl$ will not dissolve in water whereas the resulting $PbC{l_2}$ will. This is due to the fact that $PbC{l_2}$ is soluble in hot water but insoluble in cold water, unlike $AgCl$ , which is insoluble in both conditions. $Pb\,\& \,Ag$ both produce white precipitate from their chlorides. However, $AgCl$ does not dissolve in hot water but $PbC{l_2}$ does.
Thus, dil.HCl solution added after this dissolved in hot water serves as the reagent that distinguishes between silver and lead salt.
Consequently, they can be separated in this manner.
Hence, option B. is the right answer.
Note: When the ionic product is greater than the solubility product, a precipitate is produced. The production of a specific precipitate typically indicates the presence of a specific ion in qualitative analyses of inorganic ions. For instance, you can say that the unknown contains chloride ions if you get silver chloride precipitate after treating it with silver nitrate.
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