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So, the AgCl and Hg₂Cl₂ remain as solid in hot water. If NH3 (aq) is added to this mixture of two precipitates, the NH3 (aq) will react with the AgCl precipitate to make Ag(NH3)2¯(aq) and Cl(aq) solution (this underlined reaction you will balance so as to get net ionic equation 4c). The Hg₂Cl₂ however will react with NH, so as to yield a grey-black precipitate. If a grey-black precipitate appears, this is a confirmatory test for Hg₂2+. The following reaction takes place between Hg₂Cl₂ and ammonia: + 2Hg₂Cl₂(s) + 4NH, (aq)à 2HgNH₂Cl(s) + 2Hg(1) + 2NHÆ 4 (aq) + 2Cl(aq)

So, the AgCl and Hg₂Cl₂ remain as solid in hot water. If NH3 (aq) is added to this mixture of two precipitates, the NH3 (aq) will react with the AgCl precipitate to make Ag(NH3)2¯(aq) and Cl(aq) solution (this underlined reaction you will balance so as to get net ionic equation 4c). The Hg₂Cl₂ however will react with NH, so as to yield a grey-black precipitate. If a grey-black precipitate appears, this is a confirmatory test for Hg₂2+. The following reaction takes place between Hg₂Cl₂ and ammonia: + 2Hg₂Cl₂(s) + 4NH, (aq)à 2HgNH₂Cl(s) + 2Hg(1) + 2NHÆ 4 (aq) + 2Cl(aq)

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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The question is The dissolution of AgCl(s) in a solution containing NH3? The image that does not say introduction is a hint
Introduction The cations of group I (Agt, Hg₂2+, Pb²+) can all be precipitated from a solution to form insoluble chloride salts. The addition of dilute HCl to a sample containing all three group I ions thus results in the formation of the solids AgCl, Hg2Cl2, PbCl2, while leaving any group II and group III ions remaining in solution. After their precipitation, the solid chlorides of group I can be washed with cold water to remove any other residual cations from groups II and III that may be present. To separate the solid chlorides of group I, distilled water is added and the mixture is gently heated. This allows for the dissolution of PbCl2, which is soluble at elevated temperatures. Quick physical separation (decantation, i.e. careful pouring) of the solution from the remaining white solid allows for PbCl2 to be separated from solid AgCl and Hg2Cl₂. To confirm for the presence of Pb²+, a solution of K2CrO4 is added, forming the yellow precipitate PbCrO4. The remaining chloride salts (AgCl and Hg2Cl2) can be separated from each other by the addition of ammonium hydroxide (NH4OH). The addition of NH4OH introduces NH3 into the solution, which forms the complex ion Ag(NH3)2. The formation of this complex ion causes AgCl(s) to dissolve since Ag(NH3)2 is highly soluble in aqueous solution. Simultaneously, the disporportionation of Hg₂2+ occurs through the addition of NH4OH, yielding a white solid (HgNH₂Cl) and a black or grey precipitate (Hg). The solution containing Agt is then removed by centrifugation and decantation, leaving the mercury containing precipitates. While the grey precipitate seen above can be considered characteristic of Hgº, the remaining supernatant containing Agt must be tested to confirm its identity. This is achieved by adding nitric acid (HNO3) to the Ag* containing supernatant, which succeeds in re-precipitating the white solid AgCl. The appearance of this white solid confirms the presence of Agt.
Transcribed Image Text:Introduction The cations of group I (Agt, Hg₂2+, Pb²+) can all be precipitated from a solution to form insoluble chloride salts. The addition of dilute HCl to a sample containing all three group I ions thus results in the formation of the solids AgCl, Hg2Cl2, PbCl2, while leaving any group II and group III ions remaining in solution. After their precipitation, the solid chlorides of group I can be washed with cold water to remove any other residual cations from groups II and III that may be present. To separate the solid chlorides of group I, distilled water is added and the mixture is gently heated. This allows for the dissolution of PbCl2, which is soluble at elevated temperatures. Quick physical separation (decantation, i.e. careful pouring) of the solution from the remaining white solid allows for PbCl2 to be separated from solid AgCl and Hg2Cl₂. To confirm for the presence of Pb²+, a solution of K2CrO4 is added, forming the yellow precipitate PbCrO4. The remaining chloride salts (AgCl and Hg2Cl2) can be separated from each other by the addition of ammonium hydroxide (NH4OH). The addition of NH4OH introduces NH3 into the solution, which forms the complex ion Ag(NH3)2. The formation of this complex ion causes AgCl(s) to dissolve since Ag(NH3)2 is highly soluble in aqueous solution. Simultaneously, the disporportionation of Hg₂2+ occurs through the addition of NH4OH, yielding a white solid (HgNH₂Cl) and a black or grey precipitate (Hg). The solution containing Agt is then removed by centrifugation and decantation, leaving the mercury containing precipitates. While the grey precipitate seen above can be considered characteristic of Hgº, the remaining supernatant containing Agt must be tested to confirm its identity. This is achieved by adding nitric acid (HNO3) to the Ag* containing supernatant, which succeeds in re-precipitating the white solid AgCl. The appearance of this white solid confirms the presence of Agt.
So, the AgCl and Hg₂Cl₂ remain as solid in hot water. If NH₂ (aq) is added to this mixture of two precipitates, the NH₂ (aq) will react with the AgCl precipitate to make Ag(NH₂)₂ (aq) and Cl(aq) solution (this underlined reaction you will balance so as to get net ionic equation 4c). The Hg₂Cl₂ however will react with NH₂ so as to yield a grey-black precipitate. If a grey-black precipitate appears, this is a confirmatory test for Hg₂+. The following reaction takes place between Hg₂Cl₂ and ammonia: 2+ + 2Hg₂Cl₂(s) + 4NH₂ (aq)à 2HgNH₂Cl(s) + 2Hg(1) + 2NH * (aq) + 2Cl(aq)
Transcribed Image Text:So, the AgCl and Hg₂Cl₂ remain as solid in hot water. If NH₂ (aq) is added to this mixture of two precipitates, the NH₂ (aq) will react with the AgCl precipitate to make Ag(NH₂)₂ (aq) and Cl(aq) solution (this underlined reaction you will balance so as to get net ionic equation 4c). The Hg₂Cl₂ however will react with NH₂ so as to yield a grey-black precipitate. If a grey-black precipitate appears, this is a confirmatory test for Hg₂+. The following reaction takes place between Hg₂Cl₂ and ammonia: 2+ + 2Hg₂Cl₂(s) + 4NH₂ (aq)à 2HgNH₂Cl(s) + 2Hg(1) + 2NH * (aq) + 2Cl(aq)
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