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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