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Chapter 16: Problem 96

Describe how you could separate the ions in each of the following groups by selective precipitation. a. \(\mathrm{Ag}^{+}, \mathrm{Mg}^{2+}, \mathrm{Cu}^{2+}\) b. \(\mathrm{Pb}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Fe}^{2+}\) c. \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\)

Short Answer

Expert verified
To separate the ions in each group by selective precipitation: a. For \(\mathrm{Ag}^{+}, \mathrm{Mg}^{2+}, \mathrm{Cu}^{2+}\): 1. Add dilute HCl, which forms \(\mathrm{AgCl}\) precipitate. Filter to separate \(\mathrm{AgCl}\). 2. Add NaOH, which forms \(\mathrm{Cu(OH)_2}\) precipitate. Filter to separate \(\mathrm{Cu(OH)_2}\) and obtain the remaining \(\mathrm{Mg}^{2+}\). b. For \(\mathrm{Pb}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Fe}^{2+}\): 1. Add NaOH, forming \(\mathrm{Pb(OH)_2}\) and \(\mathrm{Fe(OH)_2}\) precipitates. Filter and save them. 2. Dissolve the saved precipitates in dilute HCl, then add K2CrO4, forming \(\mathrm{PbCrO_4}\) precipitate. Filter to separate \(\mathrm{PbCrO_4}\) and obtain the remaining \(\mathrm{Fe}^{2+}\). c. For \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\): 1. Add dilute HCl, forming \(\mathrm{PbCl_2}\) precipitate. Filter to separate \(\mathrm{PbCl_2}\) and obtain the remaining \(\mathrm{Bi}^{3+}\).

Step by step solution

01

Precipitate \(\mathrm{Ag}^{+}\)

Add a few drops of dilute hydrochloric acid (HCl) to the solution containing the ions. \(\mathrm{Ag}^{+}\) will form a precipitate with chloride ions as \(\mathrm{AgCl}\) while \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cu}^{2+}\) will not: \[\mathrm{Ag}^{+} + \mathrm{Cl}^{-} \rightarrow \mathrm{AgCl (s)}\] Then, filter the precipitate to separate the \(\mathrm{AgCl}\) from the other ions.
02

Precipitate \(\mathrm{Cu}^{2+}\)

Add a few drops of a sodium hydroxide (NaOH) solution to the remaining filtrate. \(\mathrm{Cu}^{2+}\) will form a blue precipitate with hydroxide ions as \(\mathrm{Cu(OH)_2}\) while \(\mathrm{Mg}^{2+}\) will not: \[\mathrm{Cu}^{2+} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Cu(OH)_2 (s)}\] Filter the precipitate, and \(\mathrm{Mg}^{2+}\) will remain in the solution. b. Separate \(\mathrm{Pb}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Fe}^{2+}\)
03

Precipitate \(\mathrm{Pb}^{2+}\) and \(\mathrm{Fe}^{2+}\)

Add a few drops of sodium hydroxide (NaOH) solution to the solution containing the ions. Both \(\mathrm{Pb}^{2+}\) and \(\mathrm{Fe}^{2+}\) will form precipitates with hydroxide ions while \(\mathrm{Ca}^{2+}\) will not: \[\mathrm{Pb}^{2+} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Pb(OH)_2 (s)}\] \[\mathrm{Fe}^{2+} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Fe(OH)_2 (s)}\] Filter the precipitates and save them for the next step.
04

Separate \(\mathrm{Pb}^{2+}\) from \(\mathrm{Fe}^{2+}\)

Dissolve the mixture of \(\mathrm{Pb(OH)_2}\) and \(\mathrm{Fe(OH)_2}\) in dilute hydrochloric acid (HCl). \(\mathrm{Pb(OH)_2}\) will form \(\mathrm{Pb}^{2+}\) and \(\mathrm{Fe(OH)_2}\) will form \(\mathrm{Fe}^{2+}\). Add a few drops of potassium chromate (K2CrO4) to this solution. \(\mathrm{Pb}^{2+}\) will form a yellow \(\mathrm{PbCrO_4}\) precipitate while \(\mathrm{Fe}^{2+}\) will not: \[\mathrm{Pb}^{2+} + \mathrm{CrO_4}^{2-} \rightarrow \mathrm{PbCrO_4 (s)}\] Filter this precipitate to separate it from the \(\mathrm{Fe}^{2+}\). c. Separate \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\)
05

Precipitate \(\mathrm{Pb}^{2+}\)

Add a few drops of dilute hydrochloric acid (HCl) to the solution containing the ions. \(\mathrm{Pb}^{2+}\) will form \(\mathrm{PbCl_2}\) precipitate while \(\mathrm{Bi}^{3+}\) will not: \[\mathrm{Pb}^{2+} + 2\mathrm{Cl}^{-} \rightarrow \mathrm{PbCl_2 (s)}\] Filter the precipitate to separate the \(\mathrm{PbCl_2}\) from \(\mathrm{Bi}^{3+}\).

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Most popular questions from this chapter

\(\mathrm{Ag}_{2} \mathrm{S}(s)\) has a larger molar solubility than CuS even though \(\mathrm{Ag}_{2} \mathrm{S}\) has the smaller \(K_{\mathrm{sp}}\) value. Explain how this is possible.

Sodium chloride is listed in the solubility rules as a soluble compound. Therefore, the \(K_{\mathrm{sp}}\) value for \(\mathrm{NaCl}\) is infinite. Is this statement true or false? Explain.

Magnesium hydroxide, \(\operatorname{Mg}(\mathrm{OH})_{2},\) is the active ingredient in the antacid TUMS and has a \(K_{\mathrm{sp}}\) value of \(8.9 \times 10^{-12}\) . If a 10.0 -g sample of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is placed in \(500.0 \mathrm{mL}\) of solution, calculate the moles of OH -ions present. Because the \(K_{\mathrm{sp}}\) value for \(\mathrm{Mg}(\mathrm{OH})_{2}\) is much less than \(1,\) not a lot solid dissolves in solution. Explain how \(\mathrm{Mg}(\mathrm{OH})_{2}\) works to neutralize large amounts of stomach acid.

The common ion effect for ionic solids (salts) is to significantly decrease the solubility of the ionic compound in water. Explain the common ion effect.

Aluminum ions react with the hydroxide ion to form the precipitate \(\mathrm{Al}(\mathrm{OH})_{3}(s),\) but can also react to form the soluble complex ion \(\mathrm{Al}(\mathrm{OH})_{4}^{-}.\) In terms of solubility, All \((\mathrm{OH})_{3}(s)\) will be more soluble in very acidic solutions as well as more soluble in very basic solutions. a. Write equations for the reactions that occur to increase the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in very acidic solutions and in very basic solutions. b. Let's study the pH dependence of the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in more detail. Show that the solubility of \(\mathrm{Al}(\mathrm{OH})_{3},\) as a function of \(\left[\mathrm{H}^{+}\right],\) obeys the equation $$S=\left[\mathrm{H}^{+}\right]^{3} K_{\mathrm{sp}} / K_{\mathrm{w}}^{3}+K K_{\mathrm{w}} /\left[\mathrm{H}^{+}\right]$$ where \(S=\) solubility \(=\left[\mathrm{Al}^{3+}\right]+\left[\mathrm{Al}(\mathrm{OH})_{4}^{-}\right]\) and \(K\) is the equilibrium constant for $$\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)$$ c. The value of \(K\) is 40.0 and \(K_{\mathrm{sp}}\) for \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(2 \times 10^{-32}\) Plot the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) in the ph range \(4-12.\)
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