Question

Identify the precipitate (if any) that forms when the following solutions are mixed, and write a balanced equation for each reaction. (a) \(\mathrm{NaCH}_{3} \mathrm{COO}\) and \(\mathrm{HCl},\) (b) \(\mathrm{KOH}\) and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2},(\mathbf{c}) \mathrm{Na}_{2} \mathrm{S}\) and \(\mathrm{CdSO}_{4}\) .

Short answer

In solution (a), there is no precipitate formation. In solution (b), Copper (II) Hydroxide (\(\mathrm{Cu}(\mathrm{OH})_2\)) precipitates, and the balanced equation is: \[\mathrm{2KOH} + \mathrm{Cu}(\mathrm{NO}_3)_2 \rightarrow \mathrm{2KNO_3} + \mathrm{Cu}(\mathrm{OH})_2\downarrow\] In solution (c), Cadmium Sulfide (\(\mathrm{CdS}\)) precipitates, and the balanced equation is: \[\mathrm{Na}_2\mathrm{S} + \mathrm{CdSO}_4 \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + \mathrm{CdS}\downarrow\]

Step-by-step solution

Identify possible products by exchanging ions

For reaction (a) of \(\mathrm{NaCH}_3\mathrm{COO}\) and \(\mathrm{HCl}\), the possible products after exchanging ions are Sodium Chloride (NaCl) and Acetic Acid (\(\mathrm{CH}_3\mathrm{COOH}\)).

Check for precipitate formation

Sodium Chloride (NaCl) and Acetic Acid (\(\mathrm{CH}_3\mathrm{COOH}\)) are both soluble in water and no precipitate will form.

Write and balance the equation (if there is a precipitate)

There is no precipitate so there's no need to write a balanced equation. ### Solution B ###

Identify possible products by exchanging ions

For reaction (b) of \(\mathrm{KOH}\) and \(\mathrm{Cu}(\mathrm{NO}_3)_2\), the possible products after exchanging ions are Potassium Nitrate (KNO3) and Copper (II) Hydroxide (\(\mathrm{Cu}(\mathrm{OH})_2\)).

Check for precipitate formation

Potassium Nitrate (KNO3) is soluble in water, but Copper (II) Hydroxide (\(\mathrm{Cu}(\mathrm{OH})_2\)) is insoluble and will precipitate.

Write and balance the equation (if there is a precipitate)

The balanced equation for this reaction is: \[\mathrm{2KOH} + \mathrm{Cu}(\mathrm{NO}_3)_2 \rightarrow \mathrm{2KNO_3} + \mathrm{Cu}(\mathrm{OH})_2\downarrow\] ### Solution C ###

Identify possible products by exchanging ions

For reaction (c) of \(\mathrm{Na}_2\mathrm{S}\) and \(\mathrm{CdSO}_4\), the possible products after exchanging ions are Sodium Sulfate (\(\mathrm{Na}_2\mathrm{SO}_4\)) and Cadmium Sulfide (\(\mathrm{CdS}\)).

Check for precipitate formation

Sodium Sulfate (\(\mathrm{Na}_2\mathrm{SO}_4\)) is soluble in water, but Cadmium Sulfide (\(\mathrm{CdS}\)) is insoluble and will precipitate.

Write and balance the equation (if there is a precipitate)

The balanced equation for this reaction is: \[\mathrm{Na}_2\mathrm{S} + \mathrm{CdSO}_4 \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + \mathrm{CdS}\downarrow\]

Key concepts

Solubility Rules

Understanding solubility rules is essential when predicting whether a precipitate will form during a chemical reaction. These rules are guidelines that help us determine the solubility of various ionic compounds in water. A substance that is soluble will dissolve in water, while an insoluble one will form a precipitate.

For example, a key rule is that all nitrates ((NO_3)−) and most salts containing the alkali metal ions (like sodium ((Na^+)) and potassium ((K^+))) are soluble, while certain hydroxides (such as (Cu(OH)_2)) and sulfides (like (CdS)), are generally insoluble. Applying these rules aids in predicting the outcomes of precipitation reactions, like the ones described in the exercise solutions.

Chemical Equations

A balanced chemical equation provides a clear representation of a chemical reaction. It shows the reactants on the left, the products on the right, and uses coefficients to ensure that the number of atoms for each element is conserved.

In the given exercise, balancing equations allows us to see that no precipitate forms in reaction (a), while in reactions (b) and (c), we identify (Cu(OH)_2) and (CdS) as precipitates, respectively. Balancing also confirms that mass and charge are balanced on both sides of the equation, a fundamental principle known as the Law of Conservation of Mass.

Ionic Compounds

An ionic compound consists of positively charged ions called cations and negatively charged ions known as anions. Ionic compounds are typically formed when metals react with nonmetals. In water, many of these compounds dissociate into their constituent ions.

This dissociation plays a crucial role in precipitation reactions. As seen in the exercise solutions, sodium acetate ((Na(CH_3COO))) dissociates into sodium ((Na^+)) and acetate ((CH_3COO^−)) ions, while hydrochloric acid ((HCl)) dissociates into hydrogen ((H^+)) and chloride ((Cl^−)) ions. The possible products are a result of the recombination of these ions.

Double Displacement Reactions

Double displacement reactions involve the exchange of ions between two compounds to form two new compounds. These reactions are also called metathesis reactions. In the context of precipitation reactions, double displacement can result in the formation of at least one insoluble product, the precipitate.

Reactions (b) and (c) from the exercise are classic examples where (K^+) ions from potassium hydroxide ((KOH)) exchange with (Cu^2+) ions from copper(II) nitrate, and (Na^+) ions from sodium sulfide ((Na_2S)) exchange with (Cd^2+) ions from cadmium sulfate ((CdSO_4)), respectively, forming the insoluble precipitates noted in the balanced equations.