Problem 84 - General Chemistry: Principles and Modern Applications

Question

To separate \(\mathrm{Fe}^{3+}\) and \(\mathrm{Ni}^{2+}\) from an aqueous solution containing both ions, with one cation forming a precipitate and the other remaining in solution, add to the solution (a) \(\mathrm{NaOH}(\mathrm{aq}) ;\) (b) \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) (c) \(\mathrm{HCl}(\mathrm{aq}) ;(\mathrm{d}) \mathrm{NH}_{3}(\mathrm{aq})\)

Step-by-Step Solution

Answer

Option (a) and (d) let \(\mathrm{Fe}^{3+}\) to form a precipitate and \(\mathrm{Ni}^{2+}\) remains in solution, while option (b) let \(\mathrm{Ni}^{2+}\) to form precipitate leaving \(\mathrm{Fe}^{3+}\) in solution. Option (c) would not be efficient for separating these ions.

Step 1: Understanding the Reactivity of Metals

It's important to note that iron (Fe) tends to form hydroxide complexes in the presence of a strong base like \(\mathrm{NaOH}(\mathrm{aq})\) and ammonia \(\mathrm{NH}_{3}(\mathrm{aq})\), whereas nickel (Ni) will remain in solution. On the other hand, in the presence of \(\mathrm{HCl}(\mathrm{aq})\), both nickel and iron will remain in solution, rendering it incapable of separating the two. As for \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\), nickel sulphide is not soluble, so it'll form a precipitate.

Step 2: Evaluate each Substance Separately

(a) If we add \(\mathrm{NaOH}(\mathrm{aq})\), iron will form a precipitate as Fe(OH)3 while nickel will remain in the solution.\n(b) If we add \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\), nickel will precipitate out as NiS, leaving iron in the solution.\n(c) Adding \(\mathrm{HCl}(\mathrm{aq})\) would make both Ni and Fe remain in solution as NiCl2 and FeCl3, thus not effective in separating them.\n(d) If we add \(\mathrm{NH}_{3}(\mathrm{aq})\), Fe will precipitate as Fe(OH)3 while Ni will remain in the solution.

Key Concepts

Chemical Precipitation

Chemical precipitation is a process used to separate ions from a solution. It involves the transformation of dissolved ions into solid particles, which can be easily removed. This technique is pivotal in separating metal ions based on their solubility characteristics.

When a compound is added to the solution, it may react with one of the ions, forming a precipitate. A precipitate is an insoluble solid that emerges from the reaction of two solutions. This is particularly common with metal ions. For instance, in our exercise, adding NaOH transforms \(\text{Fe}^{3+}\) into the insoluble \(\text{Fe(OH)}_{3}\) while \(\text{Ni}^{2+}\) stays dissolved.

• This process relies heavily on the differing solubility rules of various compounds.
• Knowing which compound forms a precipitate is key for effective separation.

By carefully selecting reagents, it's possible to isolate specific ions, making chemical precipitation a valuable tool in chemistry.

Reactivity of Metals

The reactivity of metals plays a crucial role in their separation from solutions. It's all about which metal ion reacts under given conditions. Metals have different tendencies to form compounds, and these reactions can be harnessed for separation.

For instance, \(\text{Fe}^{3+}\) reacts with bases like NaOH and NH\(\text{3}\), changing into hydroxides, which are typically less soluble in water. This reactivity results in the formation of a precipitate. On the other hand, nickel ions remain largely unreactive under such conditions, staying in the solution.

• Understanding metal reactivity can help predict the outcomes of adding certain reagents.
• Knowing which metals will react with which agents is critical for successful ion separation.

In our example, choosing the right reagent resulted in the successful separation of iron and nickel using their reactivity differences.

Coordination Chemistry

Coordination chemistry delves into the behavior of chemical species formed between metal ions and ligands. Ligands are molecules that can donate electron pairs to metal ions, creating stable complexes.

This field is essential in understanding how metals behave in solutions. It explains why certain ions form precipitates while others stay dissolved. Coordination chemistry comes into play when metals form complex ions in the presence of ligands such as ammonia (NH\(\text{3}\)). \(\text{Fe}^{3+}\) in our scenario forms a complex with NH\(\text{3}\), resulting in precipitation as \(\text{Fe(OH)}_{3}\).

• The type and strength of the ligand affect whether a metal will remain in solution or precipitate out.
• Coordination compounds can significantly influence the solubility and stability of metal ions in solutions.

By understanding coordination chemistry, we can manipulate conditions to either dissolve or precipitate specific metal ions.

Aqueous Solutions

Aqueous solutions consist of solutes dissolved in water, the universal solvent. Many reactions involving metal ions occur in such solutions due to water's ability to dissolve various substances.

Water's polar nature allows it to effectively dissolve ionic substances, causing ions to become dispersed within the solution. Metal ions like \(\text{Fe}^{3+}\) and \(\text{Ni}^{2+}\) can either stay dissolved or react depending on the added reagents. In our exercise, different solutions like NaOH, H\(\text{2}\)S, or HCl were added to manipulate these metal ions' states.

• Aqueous solutions facilitate the mobility and interaction of ions, crucial for precipitation reactions.
• The pH and other properties of the solution influence the state of dissolved ions.

Because water is such a versatile solvent, it is the medium where many separation techniques, like the one discussed, are performed.