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Chapter 23: Problem 84

To separate \(\mathrm{Fe}^{3+}\) and \(\mathrm{Ni}^{2+}\) from an aqueous solution containing both ions, with one cation forming a precipitate and the other remaining in solution, add to the solution (a) \(\mathrm{NaOH}(\mathrm{aq}) ;\) (b) \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) (c) \(\mathrm{HCl}(\mathrm{aq}) ;(\mathrm{d}) \mathrm{NH}_{3}(\mathrm{aq})\)

Short Answer

Expert verified
Option (a) and (d) let \(\mathrm{Fe}^{3+}\) to form a precipitate and \(\mathrm{Ni}^{2+}\) remains in solution, while option (b) let \(\mathrm{Ni}^{2+}\) to form precipitate leaving \(\mathrm{Fe}^{3+}\) in solution. Option (c) would not be efficient for separating these ions.

Step by step solution

01

Understanding the Reactivity of Metals

It's important to note that iron (Fe) tends to form hydroxide complexes in the presence of a strong base like \(\mathrm{NaOH}(\mathrm{aq})\) and ammonia \(\mathrm{NH}_{3}(\mathrm{aq})\), whereas nickel (Ni) will remain in solution. On the other hand, in the presence of \(\mathrm{HCl}(\mathrm{aq})\), both nickel and iron will remain in solution, rendering it incapable of separating the two. As for \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\), nickel sulphide is not soluble, so it'll form a precipitate.
02

Evaluate each Substance Separately

(a) If we add \(\mathrm{NaOH}(\mathrm{aq})\), iron will form a precipitate as Fe(OH)3 while nickel will remain in the solution.\n(b) If we add \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\), nickel will precipitate out as NiS, leaving iron in the solution.\n(c) Adding \(\mathrm{HCl}(\mathrm{aq})\) would make both Ni and Fe remain in solution as NiCl2 and FeCl3, thus not effective in separating them.\n(d) If we add \(\mathrm{NH}_{3}(\mathrm{aq})\), Fe will precipitate as Fe(OH)3 while Ni will remain in the solution.

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Most popular questions from this chapter

The text notes that in small quantities, zinc is an essential element (though it is toxic in higher concentrations). Tin is considered to be a toxic metal. Can you think of reasons why, for food storage, tinplate instead of galvanized iron is used in cans?

Write balanced equations for the following reactions described in the chapter. (a) \(\operatorname{Sc}(\text { l) is produced by the electrolysis of } \mathrm{Sc}_{2} \mathrm{O}_{3}\) dis solved in \(\mathrm{Na}_{3} \mathrm{ScF}_{6}(1)\) (b) Cr(s) reacts with HCl(aq) to produce a blue solution containing \(\mathrm{Cr}^{2+}(\mathrm{aq})\) (c) \(\mathrm{Cr}^{2+}(\text { aq })\) is readily oxidized by \(\mathrm{O}_{2}(\mathrm{g})\) to \(\mathrm{Cr}^{3+}(\mathrm{aq})\) (d) \(\mathrm{Ag}(\mathrm{s})\) reacts with concentrated \(\mathrm{HNO}_{3}(\mathrm{aq}),\) and \(\mathrm{NO}_{2}(\mathrm{g})\) is evolved.

Of the following, the two solids that will liberate \(\mathrm{Cl}_{2}(\mathrm{g})\) when heated with \(\mathrm{HCl}(\mathrm{aq})\) are \((\mathrm{a}) \mathrm{NaCl}(\mathrm{s})\) (b) \(\mathrm{ZnCl}_{2}(\mathrm{s}) ;(\mathrm{c}) \mathrm{MnO}_{2}(\mathrm{s}) ;(\mathrm{d}) \mathrm{CuO}(\mathrm{s}) ;(\mathrm{e}) \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{s})\) (f) \(\mathrm{NaOH}(\mathrm{s})\)

At \(400^{\circ} \mathrm{C}, \Delta G^{\circ}=-25 \mathrm{kJ}\) for the reaction \(2 \mathrm{Hg}(1)+\) \(\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{HgO}(\mathrm{s}) .\) If a sample of \(\mathrm{HgO}(\mathrm{s})\) is heated to \(400^{\circ} \mathrm{C},\) what will be the equilibrium partial pressure of \(\mathrm{O}_{2}(\mathrm{g}) ?\)

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