Question

Lanthanum metal forms cations with a charge of $3+.$ Consider the following observations about the chemistry of lanthanum: When lanthanum metal is exposed to air, a white solid (compound $\mathrm{A}$ ) is formed that contains lanthanum and one other element. When lanthanum metal is added to water, gas bubbles are observed and a different white solid (compound $\mathrm{B}$ ) is formed. Both $\mathrm{A}$ and $\mathrm{B}$ dissolve in hydrochloric acid to give a clear solution. When either of these solutions is evaporated, a soluble white solid (compound C) remains. If compound $\mathrm{C}$ is dissolved in water and sulfuric acid is added, a white precipitate (compound D) forms. (a) Propose identities for the substances $\mathrm{A},\mathrm{B},\mathrm{C},$ and $\mathrm{D}.(\mathbf{b})$ Write net ionic equations for all the reactions described. (c) Based on the preceding observations, what can be said about the position of lanthanum in the activity series (Table 4.5$)?$

Short answer

Compound A is Lanthanum Oxide (La2O3), compound B is Lanthanum Hydroxide (La(OH)3), compound C is Lanthanum Chloride (LaCl3), and compound D is Lanthanum Sulfate (La2(SO4)3). The net ionic equations for the reactions are: 1. $4\mathrm{La}(\mathrm{s})+3{\mathrm{O}}_2(\mathrm{g})\to 2{\mathrm{La}}_2{\mathrm{O}}_3(\mathrm{s})$ 2. $2\mathrm{La}(\mathrm{s})+6{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 2\mathrm{La}(\mathrm{OH}{)}_3(\mathrm{s})+3{\mathrm{H}}_2(\mathrm{g})$ 3. ${\mathrm{La}}_2{\mathrm{O}}_3(\mathrm{s})+6\mathrm{HCl}(\mathrm{aq})\to 2{\mathrm{LaCl}}_3(\mathrm{aq})+3{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$ and $\mathrm{La}(\mathrm{OH}{)}_3(\mathrm{s})+3\mathrm{HCl}(\mathrm{aq})\to {\mathrm{LaCl}}_3(\mathrm{aq})+3{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$ 4. $3{\mathrm{LaCl}}_3(\mathrm{aq})+2{\mathrm{H}}_2{\mathrm{SO}}_4(\mathrm{aq})\to {\mathrm{La}}_2({\mathrm{SO}}_4{)}_3(\mathrm{s})+6\mathrm{HCl}(\mathrm{aq})$ Lanthanum's position in the activity series is above hydrogen and below the alkali metals and alkaline earth metals, as it reacts with water but does not form a superoxide like alkali metals.

Step-by-step solution

Identifying compound A

When lanthanum metal is exposed to air, it reacts with the oxygen in the air to form a white solid compound A. Since lanthanum forms cations with a charge of $3+$, the oxide of the lanthanum would be ${\mathrm{La}}_2{\mathrm{O}}_3$. So compound A is Lanthanum Oxide (La2O3).

Identifying compound B

When lanthanum metal is added to water, gas bubbles are observed, and a different white solid compound B is formed. The gas bubbles suggest the release of hydrogen gas. Lanthanum forms cations with a charge of $3+$, so when it reacts with water, the product would be $\mathrm{La}(\mathrm{OH}{)}_3$. Compound B is Lanthanum Hydroxide (La(OH)3).

Identifying compound C

Both compound A and B dissolve in hydrochloric acid to give a clear solution. When this solution is evaporated, a soluble white solid (compound C) remains. Since lanthanum forms cations with a charge of $3+$ and hydrochloric acid contains chloride ions (Cl-), compound C would be Lanthanum Chloride (LaCl3).

Identifying compound D

When compound C is dissolved in water and sulfuric acid is added, a white precipitate (compound D) forms. Lanthanum forms cations with a charge of $3+$, and sulfuric acid contains sulfate ions (SO4^2-). Following solubility rules for common ionic compounds, a precipitate is formed when lanthanum sulfate is produced. Compound D is Lanthanum Sulfate (La2(SO4)3).

Writing net ionic equations

(a) Lanthanum metal reacts with oxygen in the air: $$4\mathrm{La}(\mathrm{s})+3{\mathrm{O}}_2(\mathrm{g})\to 2{\mathrm{La}}_2{\mathrm{O}}_3(\mathrm{s})$$ (b) Lanthanum metal reacts with water: $$2\mathrm{La}(\mathrm{s})+6{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 2\mathrm{La}(\mathrm{OH}{)}_3(\mathrm{s})+3{\mathrm{H}}_2(\mathrm{g})$$ (c) Lanthanum oxide or hydroxide reacts with hydrochloric acid: $${\mathrm{La}}_2{\mathrm{O}}_3(\mathrm{s})+6\mathrm{HCl}(\mathrm{aq})\to 2{\mathrm{LaCl}}_3(\mathrm{aq})+3{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$$ $$\mathrm{La}(\mathrm{OH}{)}_3(\mathrm{s})+3\mathrm{HCl}(\mathrm{aq})\to {\mathrm{LaCl}}_3(\mathrm{aq})+3{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$$ (d) Lanthanum chloride reacts with sulfuric acid: $$3{\mathrm{LaCl}}_3(\mathrm{aq})+2{\mathrm{H}}_2{\mathrm{SO}}_4(\mathrm{aq})\to {\mathrm{La}}_2({\mathrm{SO}}_4{)}_3(\mathrm{s})+6\mathrm{HCl}(\mathrm{aq})$$

Determining the position of lanthanum in the activity series

Lanthanum reacts with water to release hydrogen gas and forms a hydroxide. This observation indicates that lanthanum is more reactive than hydrogen. Based on the activity series (Table 4.5), lanthanum would be placed above hydrogen and below the alkali metals and alkaline earth metals, since it reacts with water but does not form a superoxide like alkali metals.

Key concepts

Lanthanum Compounds

Lanthanum is a chemical element with the symbol La and atomic number 57. It is a soft, ductile, silvery-white metal that tarnishes rapidly when exposed to air and is soft enough to be cut with a knife. Lanthanum, like other rare-earth elements, is known for forming a wide range of compounds, where it predominantly adopts a charge. Understanding lanthanum chemistry involves exploring its various compounds, commonly recognized by their distinct chemical properties.
Compound A in our exercise is identified as Lanthanum Oxide ($\mathrm{L}{\mathrm{a}}_2{\mathrm{O}}_3$. This compound is formed when lanthanum metal reacts with oxygen present in the air. It's indicative of lanthanum's reactivity towards oxygen, forming a stable ionic compound with a (+3) oxidation state for lanthanum.
In contrast, Compound B's formation upon reaction with water to produce Lanthanum Hydroxide ($\mathrm{L}\mathrm{a}(\mathrm{O}\mathrm{H}{)}_3$) and hydrogen gas demonstrates that lanthanum can also react with water, an important characteristic when considering its applications and handling precautions.
Compound C is Lanthanum Chloride ($\mathrm{L}\mathrm{a}\mathrm{C}{\mathrm{l}}_3$), a product of dissolving lanthanum oxide or hydroxide in hydrochloric acid. This compound's solubility in water and reaction with sulfuric acid to form a white precipitate, Lanthanum Sulfate ($\mathrm{L}{\mathrm{a}}_2(\mathrm{S}{\mathrm{O}}_4{)}_3$)—Compound D—show lanthanum's ability to interact with various acids, highlighting its diverse chemistry.

Net Ionic Equations

Net ionic equations provide a streamlined representation of chemical reactions by focusing only on the species that actually change during the reaction. When writing net ionic equations, spectators ions—ions that do not participate in the reaction—are omitted, leaving only the ions and molecules directly involved in the reaction process.
In our exercise, the net ionic equations are essential for displaying the chemical changes occurring with lanthanum compounds. When lanthanum oxide or hydroxide reacts with hydrochloric acid, the net ionic equations exclude the chloride ions ($\mathrm{C}{\mathrm{l}}^{-}$) and water molecules since they remain unchanged, showcasing the transformation of the lanthanum species.

Important Points in Net Ionic Equations

• Write balanced equations for the reactants and products.
• Identify and remove spectator ions from each side of the chemical equation.
• Ensure charges and atoms are balanced in the final net equation.

Understanding these equations underlines the significance of charge and the conservation of mass in chemical reactions. These equations are not simply academic exercises; they are crucial for predicting the outcome of reactions, especially when dealing with unfamiliar or complex species like lanthanum compounds.

Activity Series

The activity series is a chart of metals listed in order of decreasing reactivity. It's a practical tool used to predict whether a metal will displace another in a single displacement reaction and is instrumental in determining the reactivity of metals with water and acids.
The position of a metal within the activity series can influence its reactivity towards other substances. For instance, metals listed above hydrogen in the series, like lanthanum, are more reactive and can displace hydrogen from water or acids, releasing hydrogen gas in the process.

Lanthanum in the Activity Series

Lanthanum's behavior, as observed in the exercise, shows that it reacts vigorously with water to form a hydroxide and release hydrogen gas. This indicates that lanthanum is more reactive than hydrogen, placing it above hydrogen in the activity series. The formation of its hydroxide rather than a superoxide or peroxide further suggests that, while highly reactive, lanthanum is not as reactive as the alkali metals, which are known to form superoxides with oxygen due to their higher placement in the series. Such insights into lanthanum's reactivity can aid in predicting its behavior in various chemical processes and ensuring safe handling practices in the laboratory.