Problem 22 - General Chemistry

Question

In this problem you need to draw two pictures of solutions in beakers at different points in time. Time zero \((t=0)\) will be the hypothetical instant at which the reactants dissolve in the solution (if they dissolve) before they react. Time after mixing \((t>0)\) will be the time required to allow sufficient interaction of the materials. For now, we assume that insoluble solids have no ions in solution and do not worry about representing the stoichiometric amounts of the dissolved ions. Here is an example: Solid \(\mathrm{NaCl}\) and solid \(\mathrm{AgNO}_{3}\) are added to a beaker containing \(250 \mathrm{~mL}\) of water.Note that we are not showing the \(\mathrm{H}_{2} \mathrm{O}\), and we are representing only the ions and solids in solution. Using the same conditions as the example (adding the solids to \(\mathrm{H}_{2} \mathrm{O}\) ), draw pictures of the following: a. Solid lead(II) nitrate and solid ammonium chloride at \(t=0\) and \(t>0\) b. \(\mathrm{FeS}(s)\) and \(\mathrm{NaNO}_{3}(s)\) at \(t=0\) and \(t>0\) c. Solid lithium iodide and solid sodium carbonate at \(t=0\) and \(t>0\)

Step-by-Step Solution

Answer

Draw ions in the solution for soluble compounds after mixing, insoluble ones remain solid.

Step 1: Initial Analysis of Reactions

We begin by identifying the expected reactions and solubility of the given compounds in water. The compounds given are: a. Lead(II) nitrate \(\mathrm{Pb(NO_3)_2}\) and ammonium chloride \(\mathrm{NH_4Cl}\): Both are soluble in water.b. Iron sulfide \(\mathrm{FeS}\) and sodium nitrate \(\mathrm{NaNO_3}\): \(\mathrm{FeS}\) is insoluble and \(\mathrm{NaNO_3}\) is soluble in water.c. Lithium iodide \(\mathrm{LiI}\) and sodium carbonate \(\mathrm{Na_2CO_3}\): Both are soluble in water.

Step 2: Drawing at Time Zero (t=0)

a. Draw the beaker with \(\mathrm{Pb(NO_3)_2}\) and \(\mathrm{NH_4Cl}\) solids settled at the bottom, since no reaction has occurred yet. b. Draw the beaker with \(\mathrm{FeS}\) settled at the bottom and \(\mathrm{NaNO_3}\) also at the bottom, segregated from the \(\mathrm{FeS}\), as \(\mathrm{FeS}\) is insoluble. c. Draw the beaker with \(\mathrm{LiI}\) and \(\mathrm{Na_2CO_3}\) solids settled at the bottom.

Step 3: Drawing After Mixing (t>0)

a. Upon mixing, \(\mathrm{Pb(NO_3)_2}\) and \(\mathrm{NH_4Cl}\) dissolve, releasing \(\mathrm{Pb^{2+}}\), \(\mathrm{NO_3^-}\), \(\mathrm{NH_4^+}\), and \(\mathrm{Cl^-}\) ions. These ions are shown dispersed in the water.b. After mixing, \(\mathrm{FeS}\) remains as a solid at the bottom, while \(\mathrm{NaNO_3}\) dissolves, releasing \(\mathrm{Na^+}\) and \(\mathrm{NO_3^-}\) ions into the solution.c. After mixing, \(\mathrm{LiI}\) and \(\mathrm{Na_2CO_3}\) dissolve, releasing \(\mathrm{Li^+}\), \(\mathrm{I^-}\), \(\mathrm{Na^+}\), and \(\mathrm{CO_3^{2-}}\) ions.

Key Concepts

Solubility

Solubility is a chemical property that determines how well a substance can dissolve in a solvent, such as water. This property is influenced by temperature, pressure, and the nature of both the solute and solvent. In our exercise, we consider several solid compounds, each with different solubility in water. Here's what happens:

• Lead(II) nitrate and ammonium chloride are both soluble, meaning they dissolve easily in water and release ions.
• Iron sulfide is insoluble in water, so it remains as a solid.
• Lithium iodide and sodium carbonate are also soluble and therefore dissolve in water.

Understanding solubility helps us predict whether a substance will dissolve in a given situation. It can aid in determining the outcomes of chemical reactions, particularly when considering which ions will be present in the solution.

Dissolution

Dissolution is the process by which a solid, liquid, or gas forms a solution in a solvent. For solids, like those in our exercise, dissolution involves breaking down the solid into individual ions or molecules, which then uniformly disperse throughout the solvent.
Dissolution is important because it affects the concentration of ions in a solution, which can influence reaction rates and equilibrium. Here's a brief breakdown from the exercise:

• When \(\mathrm{Pb(NO_3)_2}\) and \(\mathrm{NH_4Cl}\) are added to water, they dissolve into \(\mathrm{Pb^{2+}}\), \(\mathrm{NO_3^-}\), \(\mathrm{NH_4^+}\), and \(\mathrm{Cl^-}\) ions.
• For iron sulfide, which doesn't dissolve, none of its components go into the solution, but sodium nitrate does dissolve into \(\mathrm{Na^+}\) and \(\mathrm{NO_3^-}\) ions.
• Lithium iodide and sodium carbonate both dissolve to release \(\mathrm{Li^+}\), \(\mathrm{I^-}\), \(\mathrm{Na^+}\), and \(\mathrm{CO_3^{2-}}\) ions into the water.

This process eventually allows for homogeneous mixtures where the solute is distributed evenly throughout the solvent.

Ion Dispersion

Ion dispersion refers to how ions, once dissociated from their respective compounds, spread throughout the solvent. This uniform distribution is crucial for reactions in solutions as it affects ion concentration and reaction progression.
In the exercise's context, after dissolution occurs at \(t>0\), we witness ion dispersion:

• For \(\mathrm{Pb(NO_3)_2}\) and \(\mathrm{NH_4Cl}\), ions such as \(\mathrm{Pb^{2+}}\) and \(\mathrm{NH_4^+}\) spread out in the water, providing a uniform ion distribution.
• In the sodium nitrate solution, \(\mathrm{Na^+}\) and \(\mathrm{NO_3^-}\) ions disperse, leaving \(\mathrm{FeS}\) at the bottom.
• For the \(\mathrm{LiI}\) and \(\mathrm{Na_2CO_3}\) mixture, ions such as \(\mathrm{Li^+}\) and \(\mathrm{CO_3^{2-}}\) distribute evenly in the solution.

Ion dispersion ensures that chemical reactions can occur throughout the entire solution rather than being concentrated in any particular area.

Precipitation Reactions

Precipitation reactions occur when two soluble ionic compounds in solution react to form an insoluble solid, called a precipitate. These reactions are indicative of solubility rules and can drastically change what's present in a solution. The exercise, though simplified, brings up scenarios where solubility is considered:

• For instance, if one were to mix solutions of \(\mathrm{NaCl}\) and \(\mathrm{AgNO_3}\), the \(\mathrm{AgCl}\) formed would be a precipitate, as it is insoluble in water.

In the provided exercise even though no precipitation reactions explicitly occur, understanding such reactions prepares students for predicting outcomes when different ionic compounds are mixed. It's an essential concept for anticipating and drawing reaction diagrams in chemistry.