# 3.1: Separation of group I cations

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Selective precipitation of a set of group I, i.e., lead(II) ($$\ce{Pb^{2+}}$$), mercury(I) ($$\ce{Hg2^{2+}}$$), and silver(I) ($$\ce{Ag^{+}}$$) is based on soluble ions rule#3 in the solubility guidelines in section 1.1 which states "Salts of chloride ($$\ce{Cl^{-}}$$), bromide ( $$\ce{Br^{-}}$$), and Iodide ( $$\ce{I^{-}}$$) are soluble, except when the cation is Lead ( $$\ce{Pb^{2+}}$$), Mercury ( $$\ce{Hg2^{2+}}$$), or Silver ( $$\ce{Ag^{+}}$$). The best source of $$\ce{Cl^{-}}$$ for precipitating group 1 cations from a test solution is $$\ce{HCl}$$, because it is a strong acid that completely dissociates in water producing $$\ce{Cl^{-}}$$ and $$\ce{H3O^{+}}$$ ions, both do not get involved in any undesired reactions under the conditions.

The $$\ce{K_{sp}}$$ expression is used to calculate $$\ce{Cl^{-}}$$ that will be sufficient to precipitate group 1 cations. The molar concentration of chloride ions i.e., [$$\ce{Cl^{-}}$$], in moles/liter in a saturated solution of the ionic compound can be calculated by rearranging their respective $$\ce{K_{sp}}$$ expression. For example, for $$\ce{AgCl}$$ solution, $$\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]$$ rearranges to:

$\left[\mathrm{Cl}^{-}\right]=K_{s p} /\left[\mathrm{Ag}^{+}\right]\nonumber$

and for $$\ce{PbCl2}$$ solution, Ksp = [Pb2+][Cl-]2 rearranges to:

$\left[C l^{-}\right]=\sqrt{K_{s p} /\left[P b^{2+}\right]}\nonumber$

The concentration of ions in the unknown sample are ~0.1 M. Plugging in 0.1M value for $$\ce{Pb^{2+}}$$ in the above equation shows that [$$\ce{Cl^{-}}$$] in a saturated solution having 0.1M $$\ce{Pb^{2+}}$$ is 1.3 x 10-2M:

$\left[C l^{-}\right]=\sqrt{K_{s p} /\left[P b^{2+}\right]}=\sqrt[2]{1.6 \times 10^{-5} / 0.1}=1.3 \times 10^{-2} \mathrm{M}\nonumber$

It means $$\ce{Cl^{-}}$$ concentration up to 1.3 x 10-2M will not cause precipitation from 0.1M $$\ce{Pb^{2+}}$$ solution. Increasing $$\ce{Cl^{-}}$$ above 0.013M will remove $$\ce{Pb^{2+}}$$ from the solution as a $$\ce{PbCl2}$$ precipitate. If 99.9% removal is desired, then 1.0 x 10-4 M $$\ce{Pb^{2+}}$$ will be left in the solution and the [$$\ce{Cl^{-}}$$] have to be raised to 0.40 M:

$\left[C l^{-}\right]=\sqrt[2]{K_{s p} /\left[P b^{2+}\right]}=\sqrt[2]{1.6 \times 10^{-5} / 1.0 \times 10^{-4}}=0.40 \mathrm{M}\nonumber$

The solubility of $$\ce{Hg2Cl2}$$ and $$\ce{AgCl}$$ is less than that of $$\ce{PbCl2}$$. So, a 0.40M $$\ce{Cl^{-}}$$ will remove more than 99.9% of $$\ce{Hg2^{2+}}$$ and $$\ce{Ag^{+}}$$ from the solution.

A sample of 20 drops of the aqueous solution is about 1 mL. In these experiments, ~15 drops of the test solution are collected in a test tube and 3 to 4 drops of 6M $$\ce{HCl}$$ are mixed with the solution. This results in about 0.9 mL total solution containing 1 to 1.3 M $$\ce{Cl^{-}}$$, which is more than twice the concentration needed to precipitate out 99.9% of group 1 cations.

A concentrated reagent (6M $$\ce{HCl}$$) is used to minimize the dilution of the test sample because the solution is centrifuged and the supernatant that is separated by decantation is used to analyze the remaining cations. A 12M $$\ce{HCl}$$ is available, but it is not used because it is a more hazardous reagent due to being more concentrated strong acid and also because if $$\ce{Cl^{-}}$$ concentration is raised to 5M or higher in the test solution, it can re-dissolve $$\ce{AgCl}$$, by forming water-soluble [$$\ce{AgCl2}$$]- complex ion.

The addition of $$\ce{HCl}$$ causes precipitation of group 1 cation as milky white suspension as shown in Figure $$\PageIndex{1}$$ and by chemical reaction equations below. The precipitates can be separated by gravity filtration, but more effective separation can be achieved by subjecting the suspension to centrifuge in a test tube. Centrifugal force forces the solid suspension to settle and pack at the bottom of the test tube from which the clear solution, called supernatant, can be poured out -a process called decantation. The precipitate is resuspended in pure water by stirring with a clean glass rod, centrifuged, and decanted again to wash out any residual impurities. The washed precipitate is used to separate and confirm the group 1 cations and the supernatant is saved for analysis of group 2, 3, 4, and 5 cations.

$\ce{ Pb^{2+}(aq) + 2Cl^{-}(aq) <=> PbCl2(s)(v)}\nonumber$

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

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

This page titled 3.1: Separation of group I cations is shared under a Public Domain license and was authored, remixed, and/or curated by Muhammad Arif Malik.