# 3.2: Separation and confirmation of individual ions in group I precipitates

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### Separation and confirmation of lead(II) ion

Solubility of $$\ce{PbCl2}$$ in water at 20 oC is about 1.1 g/100 mL, which is significantly higher than 1.9 x 10-4 g/100 mL for $$\ce{AgCl}$$ and 3.2 x 10-5 g/100 mL for $$\ce{Hg2Cl2}$$. Further, the solubility of $$\ce{PbCl2}$$ increases three-fold to about 3.2 g/100 mL in boiling water at 100 oC, while solubility $$\ce{AgCl}$$ and $$\ce{Hg2Cl2}$$ remain negligible. A 15 drops sample that is used to precipitate out group I cations corresponds to about 0.75 mL, which based on the molar mass of $$\ce{PbCl2}$$ is 278.1 g and the concentration of each ion ~0.1M, contains about 0.02 g of $$\ce{PbCl2}$$ precipitate. This 0.02 g of $$\ce{PbCl2}$$ requires ~0.6 mL of heated water for dissolution. The precipitated is re-suspended in ~2 mL water and heated in a boiling water bath to selectively dissolve $$\ce{PbCl2}$$, leaving any $$\ce{AgCl}$$ and $$\ce{Hg2Cl2}$$ almost undissolved, as shown in Figure $$\PageIndex{1}$$.

$\ce{ PbCl2 (s) <=>[Hot~water] Pb^{2+}(aq) + 2Cl^{-}(aq)}\nonumber$

The heated suspension is filtered using a heated gravity filtration set up to separate the residue comprising of $$\ce{AgCl}$$ and $$\ce{Hg2Cl2}$$ from filtrate containing dissolved $$\ce{PbCl2}$$.

The solubility of $$\ce{PbCl2}$$ is three times less at room temperature than in boiling water. Therefore, the 2 mL filtrate is cooled to room temperature to crystalize out $$\ce{PbCl2}$$ :

$\ce{Pb^{2+}(aq) + 2Cl^{-}(aq) <=>[Cold~water] PbCl2(s)}\nonumber$

If $$\ce{PbCl2}$$ crystals are observed in the filtrate upon cooling to room temperature, it is a confirmation of $$\ce{PbCl2}$$ in the test solution. If $$\ce{PbCl2}$$ concentration is low in the filtrate, the crystals may not form upon cooling. Few drops of 5M $$\ce{HCl}$$ are mixed with the filtrate to force the crystal formation based on the common ion effect of Cl- in the reactants. The formation of $$\ce{PbCl2}$$ crystals confirms $$\ce{Pb^{2+}}$$ as shown in Figure $$\PageIndex{2}$$, and no crystal formation at this stage confirms that $$\ce{Pb^{2+}}$$ was absent in the test solution.

### Separating mercury(I) ion from silver(I) ion and confirming mercury(I) ion

The residue left after filtering out $$\ce{Pb^{2+}}$$ in hot water, is washed further with 10 mL of hot water to washout residual $$\ce{PbCl2}$$. Then 2 mL of 6M aqueous $$\ce{NH3}$$ solution is passed through the residue drop by drop. Aqueous $$\ce{NH3}$$ dissolves $$\ce{AgCl}$$ precipitate by forming water soluble complex ion $$\ce{[Ag(NH3)2(aq)]^+}$$ through following series of reactions:

$\ce{AgCl(s) <=> Ag^{+}(aq) + Cl^{-}(aq)}\quad K_f = 1.8\times10^{-10}\nonumber$

$\ce{Ag^{+}(aq) + 2NH3(aq) <=> Ag(NH3)2^{+}(aq)}\quad K_f = 1.7\times10^7\nonumber$

$\text{Overall reaction:}~\ce{AgCl(aq) + 2NH3(aq) <=> Ag(NH3)2^{+}(aq) + Cl^{-}(aq)}\quad K = 3.0\times10^{-3}\nonumber$

The 2 mL filtrate is collected in a separate test tube for confirmation of $$\ce{Ag^+}$$ ion. Although $$\ce{Hg2Cl2}$$ precipitate is insoluble in water, it does slightly dissociate like all ionic compounds. The $$\ce{Hg2^{2+}}$$ ions undergo auto-oxidation or disproportionation reaction producing black Hg liquid and $$\ce{Hg2^{2+}}$$ ions. The $$\ce{Hg2^{2+}}$$ ions react with $$\ce{NH3}$$ and $$\ce{Cl^-}$$ forming white water-insoluble $$\ce{HgNH2Cl}$$ precipitate through the following series of reactions:

$\ce{Hg2Cl2(s) <=> Hg2^{2+}(aq) + 2Cl^{-}(aq)}\nonumber$

$\ce{Hg2^{2+}(aq) <=> Hg(l) + Hg^{2+}(aq)}\nonumber$

$\ce{Hg^{2+}(aq) + 2NH3(aq) + Cl^{-}(aq) <=> HgNH2Cl(s) + NH4^{+}(aq)}\nonumber$

$\text{Overall reaction:}\ce{~Hg2Cl2(s, white) + 2NH3(aq) <=> HgNH2Cl(s, white) + NH4^{+}(aq) + Cl^{-}(aq) + Hg(l, black)}\nonumber$

A mixture of white solid $$\ce{HgNH2Cl}$$ and black liquid Hg appears gray in color. Turning of white $$\ce{Hg2Cl2}$$ precipitate to grayish color upon addition of 6M $$\ce{NH3}$$ solution drops confirms $$\ce{Hg2^{2+}}$$ ions are present in the test solution as shown in Figure $$\PageIndex{3}$$. If the white precipitate redissolves leaving behind no grayish residue, it means the precipitate was $$\ce{AgCl}$$ and $$\ce{Hg2^{2+}}$$ were absent in the test solution.

### Confirming silver(I) ion

Although water-soluble complex ion $$\ce{[Ag(NH3)2(aq)]^+}$$ is quite stable, it does slightly decompose into $$\ce{Ag^+}$$ and $$\ce{NH3(aq)}$$. The excess $$\ce{NH3}$$ added to dissolve $$\ce{AgCl}$$ precipitate and the that produced by dissociation of $$\ce{[Ag(NH3)2(aq)]^+}$$ is removed by making the solution acidic by adding 6M $$\ce{HNO3}$$. The $$\ce{Cl^-}$$ formed from the dissolution of $$\ce{AgCl}$$ precipitate in the earlier reactions is still present in the medium. Decomposition of $$\ce{[Ag(NH3)2(aq)]^+}$$ in the acidic medium produces enough $$\ce{Ag^+}$$ ions to re-form white $$\ce{AgCl}$$ precipitate by the following series of equilibrium reactions.

$\ce{[Ag(NH3)2]^{+}(aq) <=> Ag^{+}(aq) + 2NH3(aq)}\nonumber$

$\ce{2NH3(aq) + 2H3O^{+}(aq) <=> 2NH4^{+}(aq) + 2H2O(l)}\nonumber$

$\ce{Ag^{+}(aq) + Cl^{-}(aq) <=> AgCl(s, white)}\nonumber$

$\text{Overall reaction:}\ce{~[Ag(NH3)2]^{+}(aq) + 2H3O^{+}(aq) + Cl^{-}(aq) <=> AgCl(s, white) + 2NH4^{+}(aq) + 2H2O(l)}\nonumber$

The formation of white $$\ce{AgCl}$$ precipitate at this stage in the acidified filtrate confirms $$\ce{Ag^+}$$ ion was present in the test solution, as shown in Figure $$\PageIndex{4}$$, and its absence confirms that $$\ce{Ag^+}$$ ion was not present in the test solution.

This page titled 3.2: Separation and confirmation of individual ions in group I precipitates is shared under a Public Domain license and was authored, remixed, and/or curated by Muhammad Arif Malik.