Question

Three solutions are mixed together to form a single solution. One contains \(0.2 \mathrm{~mol} \mathrm{~Pb}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\), the second contains \(0.1 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{~S}\), and the third contains \(0.1 \mathrm{~mol}\) \(\mathrm{CaCl}_{2}\). (a) Write the net ionic equations for the precipitation reaction or reactions that occur. (b) What are the spectator ions in the solution?

Short answer

The net ionic equations for the precipitation reactions are: a) \(Pb^{2+} + S^{2-} \rightarrow PbS \downarrow\) b) \(Ca^{2+} + S^{2-} \rightarrow CaS \downarrow\) The spectator ions in the solution are \(CH_{3}COO^{-}\) (acetate ion), \(Na^{+}\) (sodium ion), and \(Cl^{-}\) (chloride ion).

Step-by-step solution

Identify possible product ions

From the given compounds, Pb(CH3COO)2, Na2S, and CaCl2, we need to determine all possible combinations of ions present in these compounds. Pb2+ is combined with CH3COO-: \(Pb(CH_{3}COO)_{2}\rightarrow Pb^{2+}+2CH_{3}COO^{-}\) Na+ is combined with S2-: \(Na_{2}S\rightarrow 2Na^{+}+S^{2-}\) Ca2+ is combined with Cl-: \(CaCl_{2}\rightarrow Ca^{2+}+2Cl^{-}\)

Use the solubility rules

Now we will use solubility rules to determine if any insoluble compounds were precipitated from the reaction of the possible product ions. 1. PbS: Lead(II) sulfide is an insoluble compound, and it will precipitate. 2. CaS: Calcium sulfide is also insoluble, and it will precipitate. 3. PbCl2: Lead(II) chloride is slightly soluble, so it may or may not precipitate. The remaining combinations can be ignored as they will result in soluble compounds. Based on the solubility rules, the two possible precipitation reactions are: (a) Lead(II) sulfide: \(Pb^{2+} + S^{2-} \rightarrow PbS↓\) (b) Calcium sulfide: \(Ca^{2+}+S^{2-}\rightarrow CaS\downarrow\)

Write the balanced chemical equation

Now, write the balanced chemical equations for the two precipitation reactions: a) \(Pb(CH_{3}COO)_{2} +Na_{2}S \rightarrow PbS \downarrow + 2NaCH_{3}COO\) b) \(CaCl_{2} + Na_{2}S \rightarrow CaS \downarrow + 2NaCl\)

Write the net ionic equations

We will now convert the balanced chemical equations into net ionic equations. First, write the total ionic equations: a) \(Pb^{2+} + 2CH_{3}COO^{-} + 2Na^{+} + S^{2-} \rightarrow PbS \downarrow + 2Na^{+} + 2 CH_{3}COO^{-}\) b) \(Ca^{2+} + 2Cl^{-} + 2Na^{+} + S^{2-} \rightarrow CaS \downarrow + 2Na^{+} + 2Cl^{-}\) Cancel out the spectator ions: a) Net ionic equation: \(Pb^{2+} + S^{2-} \rightarrow PbS \downarrow\) b) Net ionic equation: \(Ca^{2+} + S^{2-} \rightarrow CaS \downarrow\)

Identify spectator ions

Spectator ions are the ions that do not undergo any changes, and they are not involved in the precipitation reaction. Based on the above equations, the spectator ions are: 1. \(CH_{3}COO^{-}\) (Acetate ion) 2. \(Na^{+}\) (Sodium ion) 3. \(Cl^{-}\) (Chloride ion)

Key concepts

Precipitation Reactions

Precipitation reactions occur when two solutions containing soluble salts are combined, resulting in the formation of an insoluble compound known as a precipitate. This type of reaction plays a crucial role in chemistry, especially when performing various experiments in aqueous solutions.

When solutions of different ionic compounds are mixed, the ions can start interacting. If this interaction leads to the formation of a compound that's not soluble in water, a solid precipitate will form.
To predict a precipitation reaction, you must first identify the ions in the solutions and check potential combinations to see if any will form an insoluble substance. For example, in the given exercise, mixing the ions \(Pb^{2+}\) with \(S^{2-}\) results in lead(II) sulfide \(PbS\), a classic case of precipitation as \(PbS\) is insoluble in water. The formation of \(CaS\) (calcium sulfide) also demonstrates this principle.
Recognizing precipitation reactions is essential because they are often used to isolate a specific component from a solution or analyze the solution's content.

Spectator Ions

Spectator ions are ions present in a solution during a chemical reaction that do not participate in the formation of the precipitate. They remain unchanged and do not form part of the net ionic equation.

Identifying spectator ions is vital in simplifying equations to better understand the main chemical reaction occurring. These ions essentially "watch" the reaction happen without actually being involved in it.
In our example, when \(Pb(CH_{3}COO)_{2}\), \(Na_{2}S\), and \(CaCl_{2}\) are mixed, ions such as \(CH_{3}COO^{-}\), \(Na^{+}\), and \(Cl^{-}\) do not change. Therefore, they are considered spectator ions. These ions exist on both the reactant and product sides and do not contribute to the formation of \(PbS\) and \(CaS\), highlighting their role as bystanders in the chemical reaction.

Solubility Rules

Solubility rules are essential guidelines that help chemists predict if a compound will dissolve in water. These rules are invaluable when determining whether an ionic compound will form a precipitate during a chemical reaction.

There are many rules, but some of the most crucial ones include:

• Most compounds containing alkali metal ions and ammonium \((NH_{4}^{+})\) are soluble.
• Nitrates \((NO_{3}^{-})\), acetates \((CH_{3}COO^{-})\), and most perchlorates are soluble.
• Chlorides, bromides, and iodides are generally soluble, except when paired with lead \((Pb^{2+})\), silver \((Ag^{+})\), or mercury \((Hg_2^{2+})\).
• Sulfates \((SO_{4}^{2-})\) are usually soluble, except with barium \((Ba^{2+})\), strontium \((Sr^{2+})\), lead \((Pb^{2+})\), and calcium \((Ca^{2+})\) among others.
• Sulfides are generally insoluble except when combined with alkali metals or ammonium.

Understanding these rules allows us to predict the formation of precipitates. In the provided exercise, applying the solubility rules confirms that \(PbS\) and \(CaS\) are insoluble, leading to the precipitation reactions detailed earlier. Knowing these guidelines makes it easier to determine the outcome of mixing different ionic solutions.