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)
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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
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.
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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