
The sample is dissolved in hot concentrated \[{{HCl}}\] and add few drops of ${{N}}{{{H}}_4}{{OH}}$, then acidified with dilute acetic acid, few drops of potassium ferrocyanide solution. The precipitate obtained is chocolate brown in color. Guess the presence of elements present.
A. ${{P}}{{{b}}^{2 + }}$
B. ${{C}}{{{u}}^{2 + }}$
C. ${{Z}}{{{n}}^{2 + }}$
D. ${{F}}{{{e}}^{2 + }}$
Answer
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Hint: Qualitative analysis is a process to identify an unknown substance. This identifies the cations and anions present in the unknown sample. It can also identify the chemical properties of the ions present in it. Cations are positively charged and anions are negatively charged.
Complete step-by-step answer:
All salts are ionic compounds and they contain both cations and anions. We can identify the cations and anions in the salt by adding certain reagents, i.e. chemicals.
There are some preliminary observations like color and nature of the substance. The color tells us about the possible cations present in the substance.
From the given options, ${{P}}{{{b}}^{2 + }}$ ions are generally white in color, ${{C}}{{{u}}^{2 + }}$ ions are blue in color, ${{Z}}{{{n}}^{2 + }}$ are greyish or metallic in color and ${{F}}{{{e}}^{2 + }}$ are blue or green in color.
Initially we are adding hot concentrated \[{{HCl}}\]. When ${{P}}{{{b}}^{2 + }}$ is added with hot concentrated \[{{HCl}}\], it forms a precipitate. Lead comes under group $1$. Other group $1$ cation is ${{A}}{{{g}}^{2 + }}$.
And after adding ${{N}}{{{H}}_4}{{OH}}$, the solution becomes alkaline which is then acidified with acetic acid. Then we are adding potassium ferrocyanide, chocolate brown color is produced. This occurs only when ${{C}}{{{u}}^{2 + }}$ is used.
The reaction of ${{C}}{{{u}}^{2 + }}$ with potassium ferrocyanide is given below:
${{C}}{{{u}}^{2 + }} + {\left[ {{{Fe}}{{\left( {{{CN}}} \right)}_6}} \right]^{4 - }} \to {{C}}{{{u}}_2}\left[ {{{Fe}}{{\left( {{{CN}}} \right)}_6}} \right]$
The product formed is copper ferrocyanide.
Hence the correct option is B.
Note: When copper reacts with ammonium hydroxide, it forms a precipitate of copper hydroxide. The precipitate is insoluble in excess sodium hydroxide. Moreover the precipitate is soluble in excess aqueous ammonia to give a deep blue solution.
Complete step-by-step answer:
All salts are ionic compounds and they contain both cations and anions. We can identify the cations and anions in the salt by adding certain reagents, i.e. chemicals.
There are some preliminary observations like color and nature of the substance. The color tells us about the possible cations present in the substance.
From the given options, ${{P}}{{{b}}^{2 + }}$ ions are generally white in color, ${{C}}{{{u}}^{2 + }}$ ions are blue in color, ${{Z}}{{{n}}^{2 + }}$ are greyish or metallic in color and ${{F}}{{{e}}^{2 + }}$ are blue or green in color.
Initially we are adding hot concentrated \[{{HCl}}\]. When ${{P}}{{{b}}^{2 + }}$ is added with hot concentrated \[{{HCl}}\], it forms a precipitate. Lead comes under group $1$. Other group $1$ cation is ${{A}}{{{g}}^{2 + }}$.
And after adding ${{N}}{{{H}}_4}{{OH}}$, the solution becomes alkaline which is then acidified with acetic acid. Then we are adding potassium ferrocyanide, chocolate brown color is produced. This occurs only when ${{C}}{{{u}}^{2 + }}$ is used.
The reaction of ${{C}}{{{u}}^{2 + }}$ with potassium ferrocyanide is given below:
${{C}}{{{u}}^{2 + }} + {\left[ {{{Fe}}{{\left( {{{CN}}} \right)}_6}} \right]^{4 - }} \to {{C}}{{{u}}_2}\left[ {{{Fe}}{{\left( {{{CN}}} \right)}_6}} \right]$
The product formed is copper ferrocyanide.
Hence the correct option is B.
Note: When copper reacts with ammonium hydroxide, it forms a precipitate of copper hydroxide. The precipitate is insoluble in excess sodium hydroxide. Moreover the precipitate is soluble in excess aqueous ammonia to give a deep blue solution.
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