Question

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

In the given solution, the first precipitate (with HCl) contains Ag⁺ as \(\mathrm{AgCl}\), and the second precipitate (with H₂SO₄) contains Sr²⁺ as \(\mathrm{SrSO}_4\). No precipitate is observed with NaOH, which means both Ni²⁺ and Mn²⁺ must be absent from the original solution.

Step-by-step solution

Determining precipitate with HCl

Addition of HCl solution would cause the following precipitate reactions: 1. \(\mathrm{Ag}^+ + \mathrm{Cl}^- \rightarrow \mathrm{AgCl}\) (precipitate) The other cations (Ni²⁺, Sr²⁺, and Mn²⁺) would not form a precipitate with HCl.

Determining precipitate with H₂SO₄

After the precipitate from step 1 is filtered off, addition of H₂SO₄ solution would cause the following precipitate reactions: 1. \(\mathrm{Sr}^{2+} + \mathrm{SO}_4^{2-} \rightarrow \mathrm{SrSO}_4\) (precipitate) In this case, only Sr²⁺ forms a precipitate with H₂SO₄. The other two remaining cations (Ni²⁺ and Mn²⁺) don't form a precipitate with H₂SO₄.

Determining precipitate with NaOH

After the precipitate from step 2 is filtered off, addition of NaOH solution would cause the following reactions: 1. \(\mathrm{Ni}^{2+} + 2\mathrm{OH}^- \rightarrow \mathrm{Ni(OH)}_2\) (precipitate) 2. \(\mathrm{Mn}^{2+} + 2\mathrm{OH}^- \rightarrow \mathrm{Mn(OH)}_2\) (precipitate) In this step, both Ni²⁺ and Mn²⁺ would form precipitates with NaOH. However, it's observed that no precipitate forms when NaOH is added. This means neither Ni²⁺ nor Mn²⁺ are present in the solution at this step.

Conclusion

Based on the results from the three steps above, we can conclude the following: 1. The first precipitate (with HCl) contains Ag⁺ as \(\mathrm{AgCl}\) 2. The second precipitate (with H₂SO₄) contains Sr²⁺ as \(\mathrm{SrSO}_4\) 3. Since no precipitate is observed with NaOH, both Ni²⁺ and Mn²⁺ must be absent from the original solution.