Problem 25 - General Chemistry: Principles and Modern Applications

Question

Write equations to represent the following observations. (a) A mixture of \(\hat{M g}(O H)_{2}(s)\) and \(Z n(O H)_{2}(s)\) is treated with \(\mathrm{NH}_{3}(\mathrm{aq}) .\) The \(\mathrm{Zn}(\mathrm{OH})_{2}\) dissolves, but the \(\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})\) is left behind. (b) When \(\mathrm{NaOH}(\mathrm{aq})\) is added to \(\mathrm{CuSO}_{4}(\mathrm{aq}),\) a pale blue precipitate forms. If \(\mathrm{NH}_{3}(\) aq) is added, the precipitate redissolves, producing a solution with an intense deep blue color. If this deep blue solution is made acidic with \(\mathrm{HNO}_{3}(\mathrm{aq}),\) the color is converted back to pale blue.

Step-by-Step Solution

Answer

The chemical reactions for the observations are: (a) \( Zn(OH)2(s) + 4NH3(aq) \rightarrow [Zn(NH3)4]^{2+}(aq) + 2OH^{-}(aq) \) and \(Mg(OH)2(s)\); (b) \(CuSO4(aq) + 2NaOH(aq) \rightarrow Cu(OH)2(s) \downarrow + Na2SO4(aq)\), \(Cu(OH)2(s) + 4NH3(aq) \rightarrow [Cu(NH3)4]^{2+}(aq) + 2OH^{-}(aq)\), and \([Cu(NH3)4]^{2+}(aq) + 6HNO3(aq) \rightarrow Cu(NO3)2(aq) + 4NO3^{-}(aq) + 6H2O(l) + 2NO2(g) \uparrow \)

Step 1: Writing Equation for Observation (a)

There are two parts here. One is Zn(OH)2 dissolving in NH3(aq) and the other is Mg(OH)2 not dissolving. The reaction of Zn(OH)2 with NH3(aq) can be written as:\[ Zn(OH)2(s) + 4NH3(aq) \rightarrow [Zn(NH3)4]^{2+}(aq) + 2OH^{-}(aq) \]The undissolved Mg(OH)2 is just expressed as:\[ Mg(OH)2(s) \] No reaction takes place with Mg(OH)2, so it remains in its original form.

Step 2: Writing Equation for Observation (b)

Similar to step 1, there are also multiple parts here. First, NaOH(aq) reacts with CuSO4(aq) to form a pale blue precipitate:\[ CuSO4(aq) + 2NaOH(aq) \rightarrow Cu(OH)2(s) \downarrow + Na2SO4(aq) \]Next, the formed Cu(OH)2 reacts with NH3(aq) and dissolves, forming a deep blue solution:\[ Cu(OH)2(s) + 4NH3(aq) \rightarrow [Cu(NH3)4]^{2+}(aq) + 2OH^{-}(aq) \]Finally, adding HNO3(aq) to the deep blue solution acidifies the solution and returns it back to pale blue color:\[ [Cu(NH3)4]^{2+}(aq) + 6HNO3(aq) \rightarrow Cu(NO3)2(aq) + 4NO3^{-}(aq) + 6H2O(l) + 2NO2(g) \uparrow \]

Key Concepts

Coordination Complexes

In chemistry, coordination complexes play a critical role due to their unique structures. These complexes consist of a central metal atom or ion bonded to surrounding molecules or ions, known as ligands. Understanding the formation of coordination complexes requires considering the nature of the metal ion and the type of ligands involved. A great example is the complex formation observed in the case of copper and ammonia.

• In the step-by-step solution, when Cu(OH) substitute2(s) reacts with NH3, it forms a complex ion, [Cu(NH3)4]²⁺. This involves the coordination of the copper ion with four ammonia molecules, where each ammonia donates a pair of electrons to form a stable complex.
• This complexation often significantly alters the properties and color of the solution, as seen in the deep blue color of the [Cu(NH3)4]²⁺ complex in comparison to the pale blue of Cu(OH)₂.

Coordination complexes are crucial to understanding various reactions and solubility phenomena. The interactions often dictate the nature and stability of compounds formed in solution, especially when transitioning between ionic and covalent states.

Solubility Equilibria

Solubility equilibria is a key concept in chemistry, revolving around the dissolution of compounds in a solvent. It represents the balance between the solid phase and the dissolved phase in a saturated solution. In this context, equilibria help us predict whether a compound will dissolve under certain conditions.

• In the provided exercise, the dissolution of Zn(OH)₂ in the presence of NH3 can be understood through solubility equilibria. Here, the formation of the [Zn(NH3)4]²⁺ complex affects the solubility, enabling Zn(OH)₂ to dissolve in an otherwise unfavorable condition.
• The equilibrium shifts as the formation of the complex lowers the concentration of Zn²⁺ ions in the solution, facilitating further dissolution of Zn(OH)₂.

Solubility equilibria are fundamental to predict whether a substance will form a precipitate or stay dissolved, making it a cornerstone in studying precipitation reactions.

Precipitation Reactions

Precipitation reactions occur when two soluble solutions react to form an insoluble solid, known as a precipitate. This classic reaction is a straightforward way to observe chemical changes visually.

• An example from our exercise occurs when NaOH is added to CuSO₄, resulting in the formation of Cu(OH)₂ as a pale blue precipitate. This indicates that the product is less soluble in solution.
• By understanding which compounds are likely to precipitate, chemists can predict the outcome of mixing different solutions.

Furthermore, precipitation reactions are essential in fields like analytical chemistry, where they are used for quantitative analysis of ions in solution. Understanding them provides insight into reaction mechanics and the factors that affect solubility.