Question

How could you use a precipitation reaction to separate each of the following pairs of cations? Write the formula for each reactant you would add, and write a balanced net ionic equation for each reaction. (a) \(\mathrm{K}^{+}\) and \(\mathrm{Hg}_{2}^{2+}\) (b) \(\mathrm{Pb}^{2+}\) and \(\mathrm{Ni}^{2+}\) (c) \(\mathrm{Ca}^{2+}\) and \(\mathrm{NH}_{4}^{+}\) (d) \(\mathrm{Fe}^{2+}\) and \(\mathrm{Ba}^{2+}\)

Short answer

Use specific anions to form insoluble compounds with target cations, separating them from soluble ions.

Step-by-step solution

Separate K+ and Hg2 2+

To separate potassium (\(\mathrm{K}^{+}\)) and mercury(I) (\(\mathrm{Hg}_2^{2+}\)) ions, use chloride ions (\(\mathrm{Cl}^{-}\)) to precipitate mercury(I) chloride (\(\mathrm{Hg}_2\mathrm{Cl}_2\)) which is insoluble in water.The balanced net ionic equation is:\[\mathrm{Hg}_2^{2+}\,(\mathrm{aq}) + 2\mathrm{Cl}^-\,(\mathrm{aq}) \rightarrow \mathrm{Hg}_2\mathrm{Cl}_2\,(\mathrm{s})\]Potassium ions stay in solution as potassium chloride (\(\mathrm{KCl}\)) is soluble.

Separate Pb2+ and Ni2+

To separate lead (\(\mathrm{Pb}^{2+}\)) and nickel (\(\mathrm{Ni}^{2+}\)) ions, add sulfate ions (\(\mathrm{SO}_4^{2-}\)) to precipitate lead(II) sulfate (\(\mathrm{PbSO}_4\)), which is insoluble in water.The balanced net ionic equation is:\[\mathrm{Pb}^{2+}\,(\mathrm{aq}) + \mathrm{SO}_4^{2-}\,(\mathrm{aq}) \rightarrow \mathrm{PbSO}_4\,(\mathrm{s})\]Nickel ions remain in the solution because nickel sulfate is soluble.

Separate Ca2+ and NH4+

To separate calcium (\(\mathrm{Ca}^{2+}\)) and ammonium (\(\mathrm{NH}_4^{+}\)) ions, add carbonate ions (\(\mathrm{CO}_3^{2-}\)) to precipitate calcium carbonate (\(\mathrm{CaCO}_3\)), which is insoluble.The balanced net ionic equation is:\[\mathrm{Ca}^{2+}\,(\mathrm{aq}) + \mathrm{CO}_3^{2-}\,(\mathrm{aq}) \rightarrow \mathrm{CaCO}_3\,(\mathrm{s})\]Ammonium ions stay in solution because ammonium carbonate is soluble.

Separate Fe2+ and Ba2+

To separate iron(II) (\(\mathrm{Fe}^{2+}\)) and barium (\(\mathrm{Ba}^{2+}\)) ions, add hydroxide ions (\(\mathrm{OH}^-\)) to precipitate iron(II) hydroxide (\(\mathrm{Fe(OH)}_2\)), which is insoluble.The balanced net ionic equation is:\[\mathrm{Fe}^{2+}\,(\mathrm{aq}) + 2\mathrm{OH}^-\,(\mathrm{aq}) \rightarrow \mathrm{Fe(OH)}_2\,(\mathrm{s})\]Barium ions remain in solution because barium hydroxide is more soluble.

Key concepts

Understanding Cations

Cations are positively charged ions that are a crucial component in understanding precipitation reactions. When a metal loses one or more electrons, it becomes a cation. Take the cation \(\mathrm{K}^+\), for example, which is formed when potassium loses one electron. Understanding the charge of a cation helps in predicting the behavior of these ions in a solution, especially their interactions with oppositely charged anions.
For instance, in the problem, various cations like mercury(I) \(\mathrm{Hg}_2^{2+}\), lead \(\mathrm{Pb}^{2+}\), calcium \(\mathrm{Ca}^{2+}\), and iron(II) \(\mathrm{Fe}^{2+}\) are to be separated through precipitation reactions by adding specific anions.
Each cation has unique chemical properties and bonds differently with different anions, leading to the formation of a solid precipitate if the compound formed is insoluble in water. In this exercise, understanding the properties of these cations helps us choose the correct anion to prompt a precipitation reaction for chemical separation.

Mastering Net Ionic Equations

Net ionic equations are simplified versions of chemical equations that focus on the ions participating in a reaction. They exclude spectator ions, which do not change during the reaction. Writing these equations requires:

• Identifying the ions involved in forming the precipitate.
• Balancing the charges and atoms.

A balanced net ionic equation shows only the ions that form the precipitate in a solid state. For example, in separating \(\mathrm{Pb}^{2+}\) and \(\mathrm{Ni}^{2+}\), the net ionic equation is \[\mathrm{Pb}^{2+}\,(\mathrm{aq}) + \mathrm{SO}_4^{2-}\,(\mathrm{aq}) \rightarrow \mathrm{PbSO}_4\,(\mathrm{s})\]. This equation highlights only the ions directly involved in the reaction and leaves out those that remain in solution, like \(\mathrm{Ni}^{2+}\).
Understanding net ionic equations helps you comprehend how chemical species interact on a molecular level, focusing on the essentials that drive the reaction.

Exploring Solubility Rules

Solubility rules are a set of guidelines used to predict whether a compound will dissolve in water. These rules are indispensable in precipitation reactions, as they help us determine if a particular combination of ions will result in an insoluble precipitate.Key solubility rules include:

• Most nitrate (\(\mathrm{NO}_3^-\)) salts are soluble.
• Most salts of sodium (\(\mathrm{Na}^+\)), potassium (\(\mathrm{K}^+\)), and ammonium (\(\mathrm{NH}_4^+\)) are soluble.
• Most chloride (\(\mathrm{Cl}^-\)) salts are soluble, except those of silver (\(\mathrm{Ag}^+\)), mercury(I) (\(\mathrm{Hg}_2^{2+}\)), and lead (\(\mathrm{Pb}^{2+}\)).
• Most sulfate (\(\mathrm{SO}_4^{2-}\)) salts are soluble, with exceptions such as barium (\(\mathrm{Ba}^{2+}\)), lead (\(\mathrm{Pb}^{2+}\)), and calcium (\(\mathrm{Ca}^{2+}\)).
• Most carbonate (\(\mathrm{CO}_3^{2-}\)) and phosphate (\(\mathrm{PO}_4^{3-}\)) salts are insoluble, except for those of sodium, potassium, and ammonium.

Using these rules, we can predict which combinations of cations and anions will produce non-dissolving compounds, thus forming a solid precipitate. In the exercise, solubility rules help us decide which anion to add to separate a specific pair of cations by forming an insoluble compound.

Chemical Separation via Precipitation

Chemical separation is a technique used to remove specific ions from a solution, often by turning them into insoluble compounds through precipitation. Precipitation reactions, therefore, present an effective method to facilitate chemical separation.To achieve this separation:

• Identify the ions you need to separate.
• Add a reagent containing the right anion to form an insoluble compound with one of the cations.

In the provided exercise, the separation of different cations involved creating insoluble precipitates:

• Using chloride ions to separate \(\mathrm{Hg}_2^{2+}\) from \(\mathrm{K}^+\).
• Utilizing sulfate ions for \(\mathrm{Pb}^{2+}\) and \(\mathrm{Ni}^{2+}\) separation.
• Adding carbonate ions to precipitate \(\mathrm{Ca}^{2+}\).
• Employing hydroxide ions for the separation of \(\mathrm{Fe}^{2+}\) from \(\mathrm{Ba}^{2+}\).

These steps help isolate the desired cation by forming a solid precipitate that can be filtered out of the solution, thus facilitating effective chemical separation.