2 part question. 1). An aqueous solution was tested for the presence of cationsb

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2 part question.

1). An aqueous solution was tested for the presence of cationsbelonging to Groups I-V. The first group reagent was added to 1 mLof the aqueous solution and any precipitate was separated from thesupernatant liquid. The second group reagent was added to thesupernatant liquid and any precipitate was separated, etc. In thisway, the four group reagents were sequentially added to thesupernatant liquid obtained from the previous separation step. Theexperimental results are:

1. dilute HCL ---> precipitated
2. H2S in 0.3 M HCl ---> no precipitate
3. H2S in NH3/NH4Cl ---> no precipitate
4. (NH4)2CO3 in NH3/NH4Cl ---> precipitate

Cations from which groups are or maybe present in this solution?Explain.

2. To prove the statement, PbCl2 is more soluble than AgCl andHg2Cl2 calculate the molar solubilities of AgCl, Hg2Cl2, and PbCl2in pure water.
Equilibria are:

AgCl (s) <--> Ag+ (aq) + Cl- (aq)     Ksp = 1.8 x 10^-10
Hg2Cl2(s) <--> Hg2 +2 (aq) + 2 Cl- (aq)    Ksp = 1.2 x 10^-18
PbCl2 (s)<--> Pb +2(aq) + 2Cl-(aq)          Ksp =1.7 x 10^-5

Please explain and show all steps, Thanks!

Explanation / Answer

IF the cations of a particular group are present, addition of thatgroup reagent should give a precipitate. From the experimentalresults, since only the fourth group reagent gives a preicpitate,fourth groups cations are present in the aqueous solution. . AgCl (s) Ag+ (aq) + Cl- (aq)     Ksp = 1.8 x 10^-10Letsolubility be 's', Ksp = 1.8 * 10^-10 = [Ag+] [Cl-]                               = s * s So solubility 's' = 1.8 * 10^-10                       = 1.34 * 10^-5 M . Hg2Cl2 (s) Hg2 +2 (aq) + 2 Cl-(aq)     Ksp = 1.2 x 10^-18 Ksp = 1.2 * 10^-18 = [Hg2+] [Cl-]^2                               = s * s2                               = s3 solubility,s = 1.1 * 10^ -6 M . PbCl2 (s) Pb +2(aq) + 2Cl-(aq)   .        Ksp = 1.7 x10^-5 = [Pb2+] [Cl-] 2                                    = s *s2 Solubility, s = 2.6 * 10^-2 M . so we can see from the three solubilities that lead chloride is themost soluble.

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