Problem 38 - General Chemistry: Principles and Modern Applications

Question

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?

Step-by-Step Solution

Answer

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 1: 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: \( \mathrm{Mg}^{2+}(\mathrm{aq}) + 2\mathrm{NaOH(aq)} \rightarrow \mathrm{Mg(OH)}_2(\mathrm{s}) + 2\mathrm{Na}^+(\mathrm{aq}) \) \( \mathrm{Cr}^{3+}(\mathrm{aq}) + 3\mathrm{NaOH(aq)} \rightarrow \mathrm{Cr(OH)}_3(\mathrm{s}) + 3\mathrm{Na}^+(\mathrm{aq}) \) \( \mathrm{Cr(OH)}_3(\mathrm{s}) + \mathrm{NaOH(aq)} \rightarrow \mathrm{[Cr(OH)}_4]^-(\mathrm{aq}) + \mathrm{Na}^+(\mathrm{aq}) \)

Step 2: 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.

Key Concepts

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, \([\mathrm{Cr(OH)}_4]^−\). 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 \([\mathrm{Cr(OH)}_4]^−\), 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.