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An aqueous solution of a mixture of two ...

An aqueous solution of a mixture of two inorganic salts when treated with dilute HCl, gave a precipitate (P) and a filtrate (Q). The precipitate (P) was found to dissolve in hot water. The filtrate (Q) remained unchanged, when treated with `H_(2)S` in a dilute mineral acid medium. However, it gave a green precipitate (R) in with `H_(2)S` in ammoniacal medium. The precipitate (R) gave a coloured solution (S), when treated with `H_(2)O_(2)` in an aqueous NaOH medium.
The coloured solution (S) contains

A

`Fe(SO_(4))_(3)`

B

`CuSO_(4)`

C

`ZnSO_(4)`

D

`Na_(2)CrO_(4)`

Text Solution

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The correct Answer is:
To solve the problem step by step, we can analyze the information provided about the inorganic salts and the reactions they undergo: ### Step 1: Identify the Precipitate (P) and Filtrate (Q) When the mixture of two inorganic salts is treated with dilute HCl, it produces a precipitate (P) and a filtrate (Q). - **Precipitate (P)**: This is likely to be lead(II) chloride (PbCl2), which is insoluble in cold water but dissolves in hot water. - **Filtrate (Q)**: This would be the remaining solution containing chromium ions (Cr³⁺), which are soluble in dilute HCl. **Hint**: Consider the solubility rules for common inorganic salts to identify which salts might precipitate and which would remain in solution. ### Step 2: Analyze the Behavior of Filtrate (Q) with H2S The filtrate (Q) remains unchanged when treated with H2S in a dilute mineral acid medium. This indicates that no precipitate forms, suggesting that chromium (Cr³⁺) does not precipitate in this condition. **Hint**: Remember that H2S is used to precipitate certain metal ions, particularly in a specific pH range. ### Step 3: React Filtrate (Q) with H2S in Ammoniacal Medium When Q is treated with H2S in an ammoniacal medium, a green precipitate (R) is formed. This precipitate is likely chromium(III) hydroxide (Cr(OH)3), which is formed in the presence of ammonia. **Hint**: Ammoniacal medium is crucial for the precipitation of certain metal hydroxides, especially for transition metals like chromium. ### Step 4: Reaction of Precipitate (R) with H2O2 in NaOH Medium The green precipitate (R) reacts with hydrogen peroxide (H2O2) in an aqueous NaOH medium to produce a colored solution (S). The reaction converts Cr(OH)3 to sodium chromate (Na2CrO4), which is yellow in color. **Hint**: The oxidation of Cr(III) to Cr(VI) is a key step here; consider the role of H2O2 as an oxidizing agent. ### Conclusion The colored solution (S) contains sodium chromate (Na2CrO4), which is yellow in color. ### Final Answer The colored solution (S) contains **Na2CrO4**.
To solve the problem step by step, we can analyze the information provided about the inorganic salts and the reactions they undergo: ### Step 1: Identify the Precipitate (P) and Filtrate (Q) When the mixture of two inorganic salts is treated with dilute HCl, it produces a precipitate (P) and a filtrate (Q). - **Precipitate (P)**: This is likely to be lead(II) chloride (PbCl2), which is insoluble in cold water but dissolves in hot water. - **Filtrate (Q)**: This would be the remaining solution containing chromium ions (Cr³⁺), which are soluble in dilute HCl. **Hint**: Consider the solubility rules for common inorganic salts to identify which salts might precipitate and which would remain in solution. ...
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