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Chapter 22: Problem 67

It is observed that a precipitate forms when a \(2.0 \mathrm{M} \mathrm{NaOH}(a q)\) solution is added dropwise to a \(0.10 \mathrm{M} \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) solution and that, on further addition of \(\mathrm{NaOH}(a q)\), the precipitate dissolves. Explain these observations using balanced chemical equations.

Short Answer

Expert verified
A precipitate of \(\mathrm{Cr(OH)_3}\) forms first, then dissolves to \(\mathrm{Cr(OH)_4^-}\) in excess \(\mathrm{NaOH}\).

Step by step solution

01

Identify Initial Reaction

When NaOH is added to the \(\mathrm{Cr(NO_3)_3(aq)}\) solution, chromium hydroxide \(\mathrm{Cr(OH)_3(s)}\) precipitate initially forms according to the reaction: \[ \mathrm{Cr^{3+}(aq) + 3OH^-(aq) \rightarrow Cr(OH)_3(s)} \] This equation shows that the addition of hydroxide ions from NaOH reacts with chromium ions to form a solid precipitate.
02

Dissolution upon Excess NaOH

With further addition of \(\mathrm{NaOH(aq)}\), the formed \(\mathrm{Cr(OH)_3(s)}\) precipitate dissolves due to the formation of soluble \(\mathrm{Cr(OH)_4^-} \) complex ions. This occurs as follows: \[ \mathrm{Cr(OH)_3(s) + OH^-(aq) \rightarrow Cr(OH)_4^-(aq)} \] The hydroxide ions in excess convert the solid precipitate into a soluble species.

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Key Concepts

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

Understanding Precipitation Reactions
Precipitation reactions occur when two aqueous solutions react to form an insoluble solid, known as a precipitate. In our scenario, when sodium hydroxide \(\mathrm{NaOH(aq)}\) is carefully added to a chromium nitrate \(\mathrm{Cr(NO_3)_3(aq)}\) solution, it introduces hydroxide ions \(\mathrm{OH^-}\) into the mixture. These hydroxide ions react with chromium ions \(\mathrm{Cr^{3+}}\), leading to the formation of a solid precipitate, chromium hydroxide \(\mathrm{Cr(OH)_3(s)}\).
The balanced chemical equation representing this reaction is:
\[ \mathrm{Cr^{3+}(aq) + 3OH^-(aq) \rightarrow Cr(OH)_3(s)} \]
Here are some key points:
  • The precipitate forms because chromium hydroxide is not soluble in water.
  • This reaction is typical for many metal salts that react with hydroxides to form solid metal hydroxides.
  • The precipitate visually indicates the formation of the product as a solid.
Complex Ion Formation and Its Effects
Once the initial precipitation reaction has occurred, you might observe the precipitate dissolving when more sodium hydroxide is added. This is due to a process called complex ion formation. In this scenario, excess hydroxide ions \(\mathrm{OH^-}\) combine with the chromium hydroxide precipitate to form a complex ion 鈥� tetrachromate(III) ion \(\mathrm{Cr(OH)_4^-}\).
The chemical equation detailing this transformation is:
\[ \mathrm{Cr(OH)_3(s) + OH^-(aq) \rightarrow Cr(OH)_4^-(aq)} \]
Some important aspects to understand about complex ion formation include:
  • The complex ions are soluble, which is why the precipitate dissolves.
  • This transformation happens when the formed precipitate has the ability to react further with ions in the solution.
  • Complex ion formation plays a critical role in various chemical processes, allowing reversible transitions between solid and dissolved states.
Exploring Chromium Hydroxide
Chromium hydroxide, \(\mathrm{Cr(OH)_3}\), is the key component in understanding the observed reactions. It is an amphoteric compound, meaning it can dissolve in both acids and bases. Initially, it appears as a solid precipitate but can transform into a soluble complex ion with additional reagents.
Key characteristics of chromium hydroxide include:
  • It is insoluble in water, which causes it to precipitate out in the presence of hydroxide ions.
  • Chromium hydroxide's ability to react with excess hydroxide to form complex ions is typical for amphoteric substances.
  • This dual characteristic makes chromium hydroxide an interesting substance in various chemical manipulations and experiments.
Understanding such behavior is fundamental to predicting the outcomes of similar chemical reactions.

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Most popular questions from this chapter

A \(100.0\) -mL sample of water from a salt lake has a chloride ion concentration of \(0.25 \mathrm{M} .\) Does \(\mathrm{AgCl}(s)\) precipitate from solution if \(5.0 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) \(\mathrm{AgNO}_{3}(a q)\) is added to the sample?

Some oil-well brines contain iodide ion. One particular brine sample had \(6.5 \mathrm{mg}\) of iodide ion per liter at \(25^{\circ} \mathrm{C}\). If equal volumes of this brine sample are mixed with a solution that is \(0.0010 \mathrm{M}\) each in lead(II) ion and silver(I) ion, which metal ion will precipitate as an iodide?

The equilibrium constant for the equation $$ \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{OH}^{-}(a q) \leftrightharpoons\left[\mathrm{Al}(\mathrm{OH})_{4}\right]^{-}(a q) $$ is \(K_{f}=40\) at \(25^{\circ} \mathrm{C}\). Calculate the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in a solution buffered at \(\mathrm{pH}=12.00\) at \(25^{\circ} \mathrm{C}\).

Excess \(\mathrm{HgI}_{2}(s)\) was equilibrated with a solution that is \(0.10 \mathrm{M}\) in \(\mathrm{KI}(a q) .\) Calculate the solubility of \(\mathrm{HgI}_{2}(s)\) in this solution at \(25^{\circ} \mathrm{C}\) given \(\mathrm{HgI}_{2}(s) \leftrightharpoons \mathrm{Hg}^{2+}(a q)+2 \mathrm{I}^{-}(a q) \quad K_{\mathrm{sp}}=2.9 \times 10^{-29} \mathrm{M}^{3}\) \(\operatorname{HgI}_{2}(s)+2 \mathrm{I}^{-}(a q) \leftrightharpoons\left[\mathrm{HgI}_{4}\right]^{2-}(a q) \quad K_{i}=0.79 \mathrm{M}^{-1}\).

Given the following data at \(25^{\circ} \mathrm{C}\) : solubility of \(\mathrm{I}_{2}(s)\) in \(\mathrm{H}_{2} \mathrm{O}(l)\) : \(0.0013 \mathrm{M}\) solubility of \(\mathrm{I}_{2}(s)\) in \(0.100 \mathrm{M} \mathrm{KI}(a q): \quad 0.051 \mathrm{M}\) calculate the equilibrium constants for the following set of equations: (1) \(\mathrm{I}_{2}(s) \leftrightharpoons \mathrm{I}_{2}(a q)\) (2) \(\mathrm{I}_{2}(s)+\mathrm{I}^{-}(a q) \leftrightharpoons \mathrm{I}_{3}^{-}(a q)\) (3) \(\mathrm{I}_{2}(a q)+\mathrm{I}^{-}(a q) \leftrightharpoons \mathrm{I}_{3}^{-}(a q)\)
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