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Chapter 23: Problem 38

What products are obtained when \(\mathrm{Mg}^{2+}(\mathrm{aq})\) and \(\mathrm{Cr}^{3+}(\mathrm{aq})\) are each treated with a limited amount of NaOH(aq)? With an excess of \(\mathrm{NaOH}(\) aq)? Why are the results different in these two cases?

Short Answer

Expert verified
The products when both \(\mathrm{Mg}^{2+}(\mathrm{aq})\) and \(\mathrm{Cr}^{3+}(\mathrm{aq})\) are treated with a limited amount of \(\mathrm{NaOH(aq)}\) are magnesium hydroxide and chromium hydroxide. With excess \(\mathrm{NaOH(aq)}\), an additional product, \(\mathrm{[Cr(OH)}_4]^{-}\), is formed. The difference in products is because the excess NaOH can react with the Cr(OH)3 precipitate to form a soluble complex ion.

Step by step solution

01

Identifying the reactions

When magnesium ions \(\mathrm{Mg}^{2+}(\mathrm{aq})\) and chromium ions \(\mathrm{Cr}^{3+}(\mathrm{aq})\) are treated with a limited amount of sodium hydroxide, \(\mathrm{NaOH(aq)}\), hydroxides of magnesium and chromium are formed, namely \(\mathrm{Mg(OH)}_2\) and \(\mathrm{Cr(OH)}_3\). However, if an excess of \(\mathrm{NaOH(aq)}\) is present, \(\mathrm{Cr(OH)}_3\) will not stay as it is, but reacts with NaOH to form a soluble complex \(\mathrm{[Cr(OH)}_4]^-\). The equations for the reactions are: \( \mathrm{Mg}^{2+}(\mathrm{aq}) + 2\mathrm{NaOH(aq)} \rightarrow \mathrm{Mg(OH)}_2(\mathrm{s}) + 2\mathrm{Na}^+(\mathrm{aq}) \) \( \mathrm{Cr}^{3+}(\mathrm{aq}) + 3\mathrm{NaOH(aq)} \rightarrow \mathrm{Cr(OH)}_3(\mathrm{s}) + 3\mathrm{Na}^+(\mathrm{aq}) \) \( \mathrm{Cr(OH)}_3(\mathrm{s}) + \mathrm{NaOH(aq)} \rightarrow \mathrm{[Cr(OH)}_4]^-(\mathrm{aq}) + \mathrm{Na}^+(\mathrm{aq}) \)
02

Explaining the difference in products

The difference in results in these two cases can be explained by the concept of limiting and excess reagents. With limited NaOH, all reagents are completely consumed to form the precipitates of Mg(OH)2 and Cr(OH)3, and no further reaction occurs. In contrast, with an excess of NaOH, there is enough NaOH to continue reacting with the Cr(OH)3 precipitate, converting it into a soluble complex ion, \(\mathrm{[Cr(OH)}_4]^-\), making the solution appear as if the chromium ions have disappeared.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Ions
Complex ions are fascinating chemical species that form when a central metal ion bonds with molecules or ions called ligands. In certain cases, some metal hydroxides can react further with excess hydroxide ions to form soluble complex ions. This is key in our original exercise. For instance, when
  • chromium(III) hydroxide, \(\mathrm{Cr(OH)}_3\), is treated with an excess of NaOH, it transforms into a soluble complex ion, \([\mathrm{Cr(OH)}_4]^鈭抃). This process is called complexation and it changes the nature of the compounds involved.
  • Complex ions are often more stable and dissolve in solution, which alters the visible properties of the reaction mixture.
As complex ions form, they can affect solubility and the chemical behavior of the species involved.
Hydroxides
Hydroxides are compounds formed from hydroxide ions (\(\mathrm{OH}^-\)) which bond with metal ions. These metal hydroxides are often insoluble and may precipitate out of solution as solids. In the reaction between metal ions like \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cr}^{3+}\) with a limited amount of NaOH,
  • we form magnesium hydroxide (\(\mathrm{Mg(OH)}_2\))
  • and chromium hydroxide (\(\mathrm{Cr(OH)}_3\)).
This occurs as the hydroxide ions combine with the metal ions, causing them to settle out of the solution. These precipitates are typically white or gelatinous solids that vary from case to case. In a chemistry lab, observing these solids helps identify certain metal ions.
Excess Reactants
An excess reactant is a reactant that remains after a chemical reaction has reached completion. When a reaction occurs with one reactant in excess, it can drive additional chemical changes. This happens in the instructed exercise when excess NaOH is introduced to the system:
  • With limited NaOH, the reaction halts at Mg(OH)2 and Cr(OH)3 formation.
  • However, adding excess NaOH initiates further reactions, converting \(\mathrm{Cr(OH)}_3\) into \([\mathrm{Cr(OH)}_4]^鈭抃), a complex ion.
The presence of an excess reactant changes the chemistry significantly. When excess NaOH is present, it not only provides more OH- ions but also ensures complete reaction for any additional steps like complex ion formation. This concept highlights the importance of reagent quantities in determining final reaction products.
Reaction Equations
Reaction equations are symbolic representations of chemical reactions. They show how reactants are transformed into products. In this specific scenario:
  • For magnesium: \(\mathrm{Mg^{2+}(aq) + 2NaOH(aq) \rightarrow Mg(OH)_2(s) + 2Na^+(aq)}\) reflects the formation of a solid precipitate through neutralization.
  • For chromium with limited NaOH: \(\mathrm{Cr^{3+}(aq) + 3NaOH(aq) \rightarrow Cr(OH)_3(s) + 3Na^+(aq)}\) conveys initial reactions resulting in chromium hydroxide precipitate.
  • With excess NaOH: \(\mathrm{Cr(OH)_3(s) + NaOH(aq) \rightarrow [Cr(OH)_4]^-(aq) + Na^+(aq)}\) indicates further complex ion formation.
These equations provide insight into the chemical transformations underlying the physical observations made during the reaction. Being comfortable with reading and writing reaction equations is crucial in chemistry as it helps predict the outcomes of chemical interactions.

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Most popular questions from this chapter

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.

Nearly all mercury(II) compounds exhibit covalent bonding. Mercury(II) chloride is a covalent molecule that dissolves in warm water. The stability of this compound is exploited in the determination of the levels of chloride ion in blood serum. Typical human blood serum levels range from 90 to \(115 \mathrm{mmol} \mathrm{L}^{-1}\) The chloride concentration is determined by titration with \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2} .\) The indicator used in the titration is diphenylcarbazone, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}=\mathrm{NCONHNHC}_{6} \mathrm{H}_{5}\) which complexes with the mercury(II) ion after all the chloride has reacted with the mercury(II). Free diphenylcarbazone is pink in solution, and when it is complexed with mercury(II), it is blue. Thus, the diphenylcarbazone acts as an indicator, changing from pink to blue when the first excess of mercury(II) appears. In an experiment, \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(\) aq) solution is standardized by titrating \(2.00 \mathrm{mL}\) of \(0.0108 \mathrm{M} \mathrm{NaCl}\) solution. It takes \(1.12 \mathrm{mL}\) of \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\) to reach the diphenylcarbazone end point. A 0.500 mL serum sample is treated with 3.50 mL water, 0.50 mL of 10\% sodium tungstate solution, and \(0.50 \mathrm{mL}\) of \(0.33 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) to precipitate proteins. After the proteins are precipitated, the sample is filtered and a \(2.00 \mathrm{mL}\) aliquot of the filtrate is titrated with \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) solution, requiring \(1.23 \mathrm{mL}\). Calculate the concentration of Cl^- Express your answer in mmol L \(^{-1}\). Does this concentration fall in the normal range?

In your own words, define the following terms: (a) domain; (b) flotation; (c) leaching; (d) amalgam.

Briefly describe each of the following ideas, phenomena, or methods: (a) lanthanide contraction; (b) zone refining; (c) basic oxygen process; (d) slag formation.

Show that under the following conditions, \(\mathrm{Ba}^{2+}(\mathrm{aq})\) can be separated from \(\mathrm{Sr}^{2+}(\mathrm{aq})\) and \(\mathrm{Ca}^{2+}(\) aq) by precipitating \(\mathrm{BaCrO}_{4}(\mathrm{s})\) with the other ions remaining in solution: $$\begin{aligned} &\left[\mathrm{Ba}^{2+}\right]=\left[\mathrm{Sr}^{2+}\right]=\left[\mathrm{Ca}^{2+}\right]=0.10 \mathrm{M}\\\ &\left[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right]=\left[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right]=1.0 \mathrm{M}\\\ &\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]=0.0010 \mathrm{M}\\\ &\mathrm{K}_{\mathrm{sp}}\left(\mathrm{BaCrO}_{4}\right)=1.2 \times 10^{-10}\\\ &K_{\mathrm{sp}}\left(\mathrm{SrCrO}_{4}\right)=2.2 \times 10^{-5} \end{aligned}$$ Use data from this and previous chapters, as necessary.
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