
When \[FeS{O_4}\] solution mixed with \[{\left( {N{H_4}} \right)_2}S{O_4}\] solution in \[1:1\] molar ratio gives the test of \[F{e^{2 + }}\] ion but \[CuS{O_4}\] solution mixed with aqueous ammonia in \[1:4\] molar ratio does not give the test of \[C{u^{2 + }}\] ion. Explain why?
Answer
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Hint:The solution which dissociates into ions releasing the corresponding ion will give a positive test of that ion and the one which does not dissociate into ions will not respond to the test for the respective ion.
Complete step by step answer:Two separate reactions are given here which lead to two different products. Let us evaluate the two reactions one by one.
When a solution of \[FeS{O_4}\] is mixed with \[{\left( {N{H_4}} \right)_2}S{O_4}\] solution a chemical reaction takes place to produce a new compound. This compound is a double salt known as Mohr’s salt. The chemical formula is\[FeS{O_4}{\left( {N{H_4}} \right)_2}S{O_4}.6{H_2}O\]. The reaction is written as:
$FeS{O_4} + {(N{H_4})_2}S{O_4} + 6{H_2}O \to FeS{O_4}{(N{H_4})_2}S{O_4}.6{H_2}O$
When a solution of \[CuS{O_4}\] is mixed with an aqueous solution of ammonia a chemical reaction takes place to produce a new complex compound. This compound is a cationic coordination complex of copper and ammonia. The chemical formula is\[\left[ {Cu{{\left( {N{H_3}} \right)}_4}S{O_4}} \right]5{H_2}O\] . The reaction is written as:
$CuS{O_4} + 4N{H_3} + 5{H_2}O \to [Cu{(N{H_3})_4}S{O_4}].5{H_2}O$
From the above reactions it is clear that the two products are completely different and will show different properties in solution. The former is a double salt and will dissociate into free ions containing \[F{e^{2 + }}\] ion. So it will respond to the test of \[F{e^{2 + }}\] .
The latter is a coordination complex compound in which the copper ion is trapped in a complex. So the \[C{u^{2 + }}\] ion is not dissociated into free ion from this complex and hence will not respond to the test of \[C{u^{2 + }}\].
Note:Double salts should not be confused with complex salts. The double salts on dissolution in water dissociates completely into free ions. The solution of complex salts remains unchanged on dissolution in water.
Complete step by step answer:Two separate reactions are given here which lead to two different products. Let us evaluate the two reactions one by one.
When a solution of \[FeS{O_4}\] is mixed with \[{\left( {N{H_4}} \right)_2}S{O_4}\] solution a chemical reaction takes place to produce a new compound. This compound is a double salt known as Mohr’s salt. The chemical formula is\[FeS{O_4}{\left( {N{H_4}} \right)_2}S{O_4}.6{H_2}O\]. The reaction is written as:
$FeS{O_4} + {(N{H_4})_2}S{O_4} + 6{H_2}O \to FeS{O_4}{(N{H_4})_2}S{O_4}.6{H_2}O$
When a solution of \[CuS{O_4}\] is mixed with an aqueous solution of ammonia a chemical reaction takes place to produce a new complex compound. This compound is a cationic coordination complex of copper and ammonia. The chemical formula is\[\left[ {Cu{{\left( {N{H_3}} \right)}_4}S{O_4}} \right]5{H_2}O\] . The reaction is written as:
$CuS{O_4} + 4N{H_3} + 5{H_2}O \to [Cu{(N{H_3})_4}S{O_4}].5{H_2}O$
From the above reactions it is clear that the two products are completely different and will show different properties in solution. The former is a double salt and will dissociate into free ions containing \[F{e^{2 + }}\] ion. So it will respond to the test of \[F{e^{2 + }}\] .
The latter is a coordination complex compound in which the copper ion is trapped in a complex. So the \[C{u^{2 + }}\] ion is not dissociated into free ion from this complex and hence will not respond to the test of \[C{u^{2 + }}\].
Note:Double salts should not be confused with complex salts. The double salts on dissolution in water dissociates completely into free ions. The solution of complex salts remains unchanged on dissolution in water.
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