Question

Give an example how each of the following insoluble ionic compounds could be produced using a precipitation reaction. Write the balanced formula equation for each reaction. a. \(\mathrm{Fe}(\mathrm{OH})_{3}(s)\) c. \(\mathrm{PbSO}_{4}(s)\) b. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\) d. \(\mathrm{BaCrO}_{4}(s)\)

Short answer

a. \[\mathrm{Fe(NO_{3})_{3}(aq) + 3NaOH(aq) \rightarrow Fe(OH)_{3}(s) + 3NaNO_{3}(aq)}\] b. \[\mathrm{Hg_{2}(NO_{3})_{2}(aq) + 2NaCl(aq) \rightarrow Hg_{2}Cl_{2}(s) + 2NaNO_{3}(aq)}\] c. \[\mathrm{Pb(NO_{3})_{2}(aq) + Na_{2}SO_{4}(aq) \rightarrow PbSO_{4}(s) + 2NaNO_{3}(aq)}\] d. \[\mathrm{Ba(NO_{3})_{2}(aq) + K_{2}CrO_{4}(aq) \rightarrow BaCrO_{4}(s) + 2KNO_{3}(aq)}\]

Step-by-step solution

a. Formation of \(\mathrm{Fe}(\mathrm{OH})_{3}(s)\)

To produce the insoluble ionic compound \(\mathrm{Fe}(\mathrm{OH})_{3}(s)\), mix solutions containing iron(III) ions and hydroxide ions. An example would be mixing \(\mathrm{Fe(NO_{3})_{3}}\)(solution) and \(\mathrm{NaOH}\)(solution). The balanced formula equation for the reaction is: \[\mathrm{Fe(NO_{3})_{3}(aq) + 3NaOH(aq) \rightarrow Fe(OH)_{3}(s) + 3NaNO_{3}(aq)}\]

b. Formation of \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\)

For producing the insoluble ionic compound \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\), mix the solutions containing mercury(I) ions, like \(\mathrm{Hg_{2}NO_{3}(s)}\) and chloride ions, like \(\mathrm{NaCl(s)}\). The balanced formula equation for the reaction is: \[\mathrm{Hg_{2}(NO_{3})_{2}(aq) + 2NaCl(aq) \rightarrow Hg_{2}Cl_{2}(s) + 2NaNO_{3}(aq)}\]

c. Formation of \(\mathrm{PbSO}_{4}(s)\)

To form the insoluble ionic compound \(\mathrm{PbSO}_{4}(s)\), mix solutions containing lead(II) ions and sulfate ions. An example would be mixing \(\mathrm{Pb(NO_{3})_{2}}\)(solution) and \(\mathrm{Na_{2}SO_{4}}\)(solution). The balanced formula equation for the reaction is: \[\mathrm{Pb(NO_{3})_{2}(aq) + Na_{2}SO_{4}(aq) \rightarrow PbSO_{4}(s) + 2NaNO_{3}(aq)}\]

d. Formation of \(\mathrm{BaCrO}_{4}(s)\)

For producing the insoluble ionic compound \(\mathrm{BaCrO}_{4}(s)\), mix the solutions containing barium ions and chromate ions. An example would be mixing \(\mathrm{Ba(NO_{3})_{2}}\)(solution) and \(\mathrm{K_{2}CrO_{4}}\)(solution). The balanced formula equation for the reaction is: \[\mathrm{Ba(NO_{3})_{2}(aq) + K_{2}CrO_{4}(aq) \rightarrow BaCrO_{4}(s) + 2KNO_{3}(aq)}\]

Key concepts

Insoluble Ionic Compounds

In the world of chemistry, insoluble ionic compounds are substances that do not dissolve significantly in water. They remain mostly as a solid, forming a precipitate that can be seen with the naked eye. This characteristic is particularly important in precipitation reactions. For a compound to be termed insoluble, the ions must strongly attract each other, overcoming the force of the water surrounding them. When an ionic compound like this is formed, it's often through a precipitation reaction. It's essential to know the solubility rules, as they help predict which ionic compounds will precipitate out of a solution. Not every mix of ions will result in a solid precipitate, so understanding these rules is crucial for predicting and explaining chemical observations.

Balanced Chemical Equations

Balanced chemical equations are fundamental in representing chemical reactions. They show the relation between reactants and products in a precise, numerical manner, ensuring the conservation of mass. Each balanced equation adheres to the law of conservation of mass, meaning the number of atoms for each element is the same on both sides of the equation. Take the reaction to produce iron(III) hydroxide: - \[\mathrm{Fe(NO_{3})_{3}(aq) + 3NaOH(aq) \rightarrow Fe(OH)_{3}(s) + 3NaNO_{3}(aq)}\]Notice how each atom's count is identical on both sides, illustrating a perfect balance. Writing balanced equations helps chemists understand the relative amounts of each substance involved and predict the quantities needed or produced.

Iron(III) Hydroxide

Iron(III) hydroxide, \(\mathrm{Fe(OH)_{3}}\), is a compound that forms a brown solid precipitate under standard conditions. In lab settings, it's typically produced by mixing a solution of \(\mathrm{Fe^{3+}}\) ions with hydroxide ions \(\mathrm{OH^{-}}\). When these ions meet, they bond to form the insoluble precipitate, \(\mathrm{Fe(OH)_{3}(s)}\). The balanced reaction equation is:- \[\mathrm{Fe(NO_{3})_{3}(aq) + 3NaOH(aq) \rightarrow Fe(OH)_{3}(s) + 3NaNO_{3}(aq)}\]This reaction clearly shows the solid iron(III) hydroxide forming as the result of this chemical interaction.

Mercury(I) Chloride

Mercury(I) chloride, \(\mathrm{Hg_{2}Cl_{2}}\), appears as a white precipitate when formed. This compound emerges when a solution containing mercury(I) ions encounters chloride ions. For example, mixing solutions of \(\mathrm{Hg_{2}(NO_{3})_{2}}\) and \(\mathrm{NaCl}\) will yield mercury(I) chloride. The balanced chemical equation for this reaction is:- \[\mathrm{Hg_{2}(NO_{3})_{2}(aq) + 2NaCl(aq) \rightarrow Hg_{2}Cl_{2}(s) + 2NaNO_{3}(aq)}\]This equation efficiently details the transformation from soluble ions to the insoluble mercury(I) chloride precipitate.

Lead(II) Sulfate

Lead(II) sulfate, \(\mathrm{PbSO_{4}}\), is a well-known insoluble ionic solid, often found as a white precipitate in solutions. This compound can be produced by reacting lead(II) ions with sulfate ions. For example, mixing solutions of \(\mathrm{Pb(NO_{3})_{2}}\) with \(\mathrm{Na_{2}SO_{4}}\) results in the formation of lead(II) sulfate.The balanced equation for this reaction is:- \[\mathrm{Pb(NO_{3})_{2}(aq) + Na_{2}SO_{4}(aq) \rightarrow PbSO_{4}(s) + 2NaNO_{3}(aq)}\]This reaction illustrates how lead(II) sulfate precipitates out as a solid when these ions combine.

Barium Chromate

Barium chromate, \(\mathrm{BaCrO_{4}}\), is known for its characteristic yellow appearance when it precipitates. This compound is formed when barium ions react with chromate ions in a solution. For instance, by combining \(\mathrm{Ba(NO_{3})_{2}}\) and \(\mathrm{K_{2}CrO_{4}}\) solutions, the insoluble barium chromate can be obtained. The balanced equation for this reaction is:- \[\mathrm{Ba(NO_{3})_{2}(aq) + K_{2}CrO_{4}(aq) \rightarrow BaCrO_{4}(s) + 2KNO_{3}(aq)}\]This reaction demonstrates the formation of barium chromate as a solid precipitate, highlighting the beauty of precipitation reactions.