Suppose you have a solution that might contain any or all of the following cations: \(\mathrm{Ni}^{2+}, \mathrm{Ag}^{+}, \mathrm{Sr}^{2+},\) and \(\mathrm{Mn}^{2+} .\) Addition of HCl solution causes a precipitate to form. After filtering off the precipitate, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is added to the resulting solution and another precipitate forms. This is filtered off, and a solution of \(\mathrm{NaOH}\) is added to the resulting solution. No precipitate is observed. Which ions are present in each of the precipitates? Which of the four ions listed above must be absent from the original solution?

Short Answer

Expert verified
The precipitate formed after adding HCl consists of \(\mathrm{AgCl}\), and the precipitate formed after adding \(\mathrm{H}_{2} \mathrm{SO}_{4}\) consists of \(\mathrm{SrSO}_{4}\). No precipitate is observed after adding \(\mathrm{NaOH}\), indicating the absence of \(\mathrm{Ni}^{2+}\) and \(\mathrm{Mn}^{2+}\) ions in the original solution. Thus, the original solution contains \(\mathrm{Ag}^{+}\) and \(\mathrm{Sr}^{2+}\) cations.

Step by step solution

01

1. Precipitate with HCl

After the addition of HCl, a precipitate forms. This indicates the presence of cations that form insoluble chloride salts. We need to check the solubility rules for each possible cation: - \(\mathrm{Ni}^{2+}\) and \(\mathrm{Mn}^{2+}\): Both nickel(II) and manganese(II) chlorides are soluble in water, so they are not responsible for the precipitate. - \(\mathrm{Ag}^{+}\): Silver chloride (AgCl) is insoluble in water, so the silver ions would form a precipitate with the chloride ions. - \(\mathrm{Sr}^{2+}\): Strontium chloride (SrCl2) is also soluble in water, so it would not form a precipitate. Thus, the precipitate formed after adding HCl consists of \(\mathrm{AgCl}\).
02

2. Precipitate with \(\mathrm{H}_{2} \mathrm{SO}_{4}\)

After the first precipitate has been filtered, we add H2SO4 to the resulting solution and observe another precipitate formation. We need to examine the solubility rules for sulfate salts containing the remaining possible cations: - \(\mathrm{Ni}^{2+}\): Nickel(II) sulfate (NiSO4) is soluble in water, so it would not form a precipitate. - \(\mathrm{Sr}^{2+}\): Strontium sulfate (SrSO4) is slightly insoluble in water, so the strontium ions would form a precipitate with the sulfate ions. - \(\mathrm{Mn}^{2+}\): Manganese(II) sulfate (MnSO4) is soluble in water, so it would not form a precipitate. Thus, the precipitate formed after adding H2SO4 consists of \(\mathrm{SrSO}_{4}\).
03

3. Precipitate with \(\mathrm{NaOH}\)

After the second precipitate has been filtered, we add NaOH to the resulting solution and observe no precipitate formation. This indicates that none of the remaining possible cations in the solution forms insoluble hydroxide salts. - \(\mathrm{Ni}^{2+}\): Nickel(II) hydroxide (Ni(OH)2) is insoluble in water. However, since no precipitate is observed, it indicates that nickel ions are not present in the original solution. - \(\mathrm{Mn}^{2+}\): Manganese(II) hydroxide (Mn(OH)2) is also insoluble in water, but as there is no precipitate formed, manganese ions are not present in the original solution. Based on the results, we conclude that the original solution contains \(\mathrm{Ag}^{+}\) and \(\mathrm{Sr}^{2+}\) cations, while \(\mathrm{Ni}^{2+}\) and \(\mathrm{Mn}^{2+}\) ions are absent.

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Most popular questions from this chapter

Which of the following ions will always be a spectator ion in a precipitation reaction? (a) \(\mathrm{Cl}^{-},(\mathbf{b}) \mathrm{NO}_{3}^{-},(\mathbf{c}) \mathrm{NH}_{4}^{+},(\mathbf{d}) \mathrm{S}^{2}\) , (e) \(\mathrm{SO}_{4}^{2-} .\) [Section 4.2\(]\)

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