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Chapter 4: Problem 93

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 first precipitate (\(\mathrm{AgCl}\)) indicates the presence of \(\mathrm{Ag}^{+}\) ions, and the second precipitate (\(\mathrm{SrSO}_{4}\)) indicates the presence of \(\mathrm{Sr}^{2+}\) ions in the original solution. Both \(\mathrm{Ni}^{2+}\) and \(\mathrm{Mn}^{2+}\) ions are absent from the original solution.

Step by step solution

01

Identify the possible precipitates with HCl

When HCl is added to the solution containing the cations, it reacts with them and possibly forms precipitates if the compounds are sparingly soluble. Check for the solubility of the resulting halides in each case: 1. \(\mathrm{Ni}^{2+}\) forms \(\mathrm{NiCl}_{2}\) which is soluble in water. 2. \(\mathrm{Ag}^{+}\) forms \(\mathrm{AgCl}\) which is insoluble in water. 3. \(\mathrm{Sr}^{2+}\) forms \(\mathrm{SrCl}_{2}\) which is soluble in water. 4. \(\mathrm{Mn}^{2+}\) forms \(\mathrm{MnCl}_{2}\) which is soluble in water. The first precipitate is \(\mathrm{AgCl}\) which indicates the presence of \(\mathrm{Ag}^{+}\) ions.
02

Identify the possible precipitates with H2SO4

When H2SO4 is added to the solution containing the remaining cations, it reacts with them and possibly forms precipitates if the compounds are sparingly soluble. Check the solubility of the resulting sulfates in each case: 1. \(\mathrm{Ni}^{2+}\) with \(\mathrm{H}_{2} \mathrm{SO}_{4}\) forms \(\mathrm{NiSO}_{4}\) which is soluble in water. 2. \(\mathrm{Sr}^{2+}\) with \(\mathrm{H}_{2} \mathrm{SO}_{4}\) forms \(\mathrm{SrSO}_{4}\) which is sparingly soluble in water. 3. \(\mathrm{Mn}^{2+}\) with \(\mathrm{H}_{2} \mathrm{SO}_{4}\) forms \(\mathrm{MnSO}_{4}\) which is soluble in water. The second precipitate is \(\mathrm{SrSO}_{4}\) which indicates the presence of \(\mathrm{Sr}^{2+}\) ions.
03

Check for precipitate formation with NaOH

Finally, NaOH is added to the solution containing the remaining cations, and no precipitate is observed. We need to check for the formation of any insoluble hydroxides with the remaining cations: 1. \(\mathrm{Ni}^{2+}\) forms \(\mathrm{Ni(OH)}_{2}\) which is sparingly soluble in water. 2. \(\mathrm{Mn}^{2+}\) forms \(\mathrm{Mn(OH)}_{2}\) which is sparingly soluble in water. Since no precipitate is observed after adding NaOH, it means that neither \(\mathrm{Ni}^{2+}\) nor \(\mathrm{Mn}^{2+}\) ions were present in the original solution.
04

Final answer

The first precipitate (\(\mathrm{AgCl}\)) indicates the presence of \(\mathrm{Ag}^{+}\) ions, and the second precipitate (\(\mathrm{SrSO}_{4}\)) indicates the presence of \(\mathrm{Sr}^{2+}\) ions in the original solution. Both \(\mathrm{Ni}^{2+}\) and \(\mathrm{Mn}^{2+}\) ions are absent from the original solution.

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Pure acetic acid, known as glacial acetic acid, is a liquid with a density of \(1.049 \mathrm{~g} / \mathrm{mL}\) at \(25^{\circ} \mathrm{C}\). Calculate the molarity of a solution of acetic acid made by dissolving \(20.00 \mathrm{~mL}\) of glacial acetic acid at \(25^{\circ} \mathrm{C}\) in enough water to make \(250.0 \mathrm{~mL}\) of solution.

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