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Assertion : A solution of AgCI in NH(4)O...

Assertion : A solution of `AgCI` in `NH_(4)OH` gives a white prrecipitate when acidified with `NHO_(3)`
Reasion : `[Ag(NH_(3))_(2)]^(o+)` decompoes in the presence of `HNO_(3)`

A

If both (A) and (B) are correct and (R ) is the correct explqanation of (A)

B

If both (A) and (B) are correct but (R ) is not the correct explqanation of (A)

C

If (A) is correct ,but (R ) is incorrect

D

If (A) is incorrect ,but (R ) is correct

Text Solution

AI Generated Solution

The correct Answer is:
To solve the given question, we need to analyze both the assertion and the reason provided. ### Step-by-Step Solution: 1. **Understanding the Assertion**: - The assertion states that a solution of AgCl in NH4OH gives a white precipitate when acidified with HNO3. - AgCl (silver chloride) is a sparingly soluble salt that can dissolve in ammonia (NH4OH) to form a complex ion. 2. **Reaction of AgCl with NH4OH**: - When AgCl is treated with NH4OH, it forms a complex ion: \[ \text{AgCl} + 2 \text{NH}_3 \rightarrow [\text{Ag(NH}_3)_2]^+ + \text{Cl}^- \] - This complex ion, \([\text{Ag(NH}_3)_2]^+\), is soluble in the solution. 3. **Acidification with HNO3**: - When the solution is acidified with HNO3 (nitric acid), the complex ion decomposes: \[ [\text{Ag(NH}_3)_2]^+ + \text{H}^+ \rightarrow \text{Ag}^+ + \text{NH}_4^+ \] - This reaction releases Ag+ ions into the solution. 4. **Formation of Precipitate**: - The released Ag+ ions then react with the chloride ions (Cl-) present in the solution: \[ \text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl} \downarrow \] - This results in the formation of a white precipitate of AgCl. 5. **Conclusion about the Assertion**: - The assertion is correct because the reaction indeed leads to the formation of a white precipitate of AgCl when the solution is acidified. 6. **Understanding the Reason**: - The reason states that \([\text{Ag(NH}_3)_2]^+\) decomposes in the presence of HNO3. - This is also correct, as we have shown that the complex ion decomposes to release Ag+ ions upon acidification. 7. **Final Conclusion**: - Both the assertion and the reason are correct, and the reason correctly explains the assertion. ### Final Answer: - The assertion is true, and the reason is true; the reason is a correct explanation of the assertion.

To solve the given question, we need to analyze both the assertion and the reason provided. ### Step-by-Step Solution: 1. **Understanding the Assertion**: - The assertion states that a solution of AgCl in NH4OH gives a white precipitate when acidified with HNO3. - AgCl (silver chloride) is a sparingly soluble salt that can dissolve in ammonia (NH4OH) to form a complex ion. ...
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Assertion: A solution of AgCl in NH_(4)OH gives a white precipitate when acidified with HNO_(3) . Reason: [Ag(NH_(3))_(2)]^(+) decomposes in the presence of HNO_(3) .

Assertion : Solubility of AgCl in NH_(3)(aq) is greater than in pure water. Reason : When AgCl dissolve in NH_(3)(aq) , complex ion [ Ag(NH_(3))_(2)^(+) ] formation takes place and solubility equilibrium of AgCl_(3) shifted in forward direction.

Assertion: [CoCl_(3)(NH_(3))_(3)] does not give white precipitate with AgNO_(3) solution. Reason: [CoCl_(3)(NH_(3))_(3)] complex is optically inactive.

Assertion : [Co(NH_(3))_(5)Br]SO_(4) gives white precipitate with barium chloride . Reason : The complex [Co(NH_(3))_(5)Br]SO_(4) dissociates in the solution to give Br^(-) and SO_(4)^(2-) .

When NH_(4)Cl is added to a solution of NH_(4)OH :

A solution when diluted with H_(2)O And boiled gives a white precipitate .On the addition of excess NH_(4)Cl & NH_(4)OH the volume of the precipitate decreases leaving behind a white gelationtious precipitate identify the precipitate which dissolves in NH_(4)OH//NH_(4)Cl :

(i) A white solid mixture of two salts containing a common cations in insoluble in water. It dissolves in dilute HCl producing some gases (with effervescence) that turn an acidified dichromate solution gren. After the gases are passed through the acidified dichromate solution, the emerging gas turns baryta water milky. (ii) On treatment with dilute HNO_(3) , the white solid gives a solution which does not directly give a precipitate with a BaCl_(2) solution but gives a white precipitate when warmed with H_(2)O_(2) and then treated with a BaCl_(2) solution. (iii) The solution of the mixture in dilute HCl, when treated with NH_(4)Cl,NH_(4)OH and an Na_(2)HPO_(4) solution, gives a white precipitate. Q. The white precipitate obtained in (ii) indicates the presence of a:

(i) A white solid mixture of two salts containing a common cations in insoluble in water. It dissolves in dilute HCl producing some gases (with effervescence) that turn an acidified dichromate solution gren. After the gases are passed through the acidified dichromate solution, the emerging gas turns baryta water milky. (ii) On treatment with dilute HNO_(3) , the white solid gives a solution which does not directly give a precipitate with a BaCl_(2) solution but gives a white precipitate when warmed with H_(2)O_(2) and then treated with a BaCl_(2) solution. (iii) The solution of the mixture in dilute HCl, when treated with NH_(4)Cl,NH_(4)OH and an Na_(2)HPO_(4) solution, gives a white precipitate. Q. The white precipitate obtained in (iii) consists of:

(i) A white solid mixture of two salts containing a common cations in insoluble in water. It dissolves in dilute HCl producing some gases (with effervescence) that turn an acidified dichromate solution gren. After the gases are passed through the acidified dichromate solution, the emerging gas turns baryta water milky. (ii) On treatment with dilute HNO_(3) , the white solid gives a solution which does not directly give a precipitate with a BaCl_(2) solution but gives a white precipitate when warmed with H_(2)O_(2) and then treated with a BaCl_(2) solution. (iii) The solution of the mixture in dilute HCl, when treated with NH_(4)Cl,NH_(4)OH and an Na_(2)HPO_(4) solution, gives a white precipitate. Q. The gases evolved in (i) are:

When HNO_(3) is converted into NH_(3) , the equivalent weight of HNO_(3) will be: