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 \(\mathrm{HCl},(\mathrm{b}) \mathrm{KOH}\) and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\), (c) \(\mathrm{Na}_{2} \mathrm{~S}\) and \(\mathrm{CdSO}_{4}\).

Short answer

(a) No precipitate forms. Balanced equation: \(\mathrm{NaCH}_{3}\mathrm{COO} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{HCH_{3}COO}\). (b) Precipitate: \(\mathrm{Cu(OH)_2}\). Balanced equation: \(\mathrm{KOH} + \mathrm{Cu(NO}_{3})_{2} \rightarrow \mathrm{KNO}_{3} + \mathrm{Cu(OH)_{2}↓}\). (c) Precipitate: \(\mathrm{CdS}\). Balanced equation: \(\mathrm{Na}_{2}\mathrm{S} + \mathrm{CdSO}_{4} \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{4} + \mathrm{CdS ↓}\).

Step-by-step solution

Precipitate Formation for (a) NaCH3COO and HCl

First, we write the full ionic equation for the reactants: \(\mathrm{NaCH}_{3}\mathrm{COO} \rightarrow \mathrm{Na}^{+} + \mathrm{CH}_{3}\mathrm{COO}^{-}\), \(\mathrm{HCl} \rightarrow \mathrm{H}^{+} + \mathrm{Cl}^{-}\). When these ions combine, we have two possible products: \(\mathrm{NaCl}\) and \(\mathrm{HCH_3COO}\) (acetic acid). Using the solubility rules, we note that \(\mathrm{NaCl}\) is soluble since Na is an alkali metal. Acetic acid is soluble since it is a weak acid. Therefore, there is no precipitate formed in this case. Then, we write the balanced equation for the reaction: \(\mathrm{NaCH}_{3}\mathrm{COO} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{HCH_{3}COO}\).

Precipitate Formation for (b) KOH and Cu(NO3)₂

First, we write the full ionic equation for the reactants: \(\mathrm{KOH} \rightarrow \mathrm{K}^{+} + \mathrm{OH}^{-}\), \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow \mathrm{Cu}^{2+} + 2\mathrm{NO}_3^-\). When these ions combine, we have two possible products: \(\mathrm{KNO}_{3}\) and \(\mathrm{Cu(OH)_2}\). Using the solubility rules, we note that \(\mathrm{KNO}_3\) is soluble since K is an alkali metal. However, \(\mathrm{Cu(OH)_2}\) is insoluble since hydroxides are generally insoluble, except when combined with alkali metals or ammonium. So, the precipitate that forms is \(\mathrm{Cu(OH)_2}\). Then, we write the balanced equation for the reaction: \(\mathrm{KOH} + \mathrm{Cu(NO}_{3})_{2} \rightarrow \mathrm{KNO}_{3} + \mathrm{Cu(OH)_{2}↓}\).

Precipitate Formation for (c) Na₂S and CdSO₄

First, we write the full ionic equation for the reactants: \(\mathrm{Na}_{2}\mathrm{S} \rightarrow 2\mathrm{Na}^{+} + \mathrm{S}^{2-}\), \(\mathrm{CdSO}_{4} \rightarrow \mathrm{Cd}^{2+} + \mathrm{SO}_4^{2-}\). When these ions combine, we have two possible products: \(\mathrm{Na_2SO_4}\) and \(\mathrm{CdS}\). Using the solubility rules, we note that \(\mathrm{Na}_2\mathrm{SO}_4\) is soluble since Na is an alkali metal. However, \(\mathrm{CdS}\) is insoluble since sulfides are generally insoluble, except when combined with alkali metals or ammonium. So, the precipitate that forms is \(\mathrm{CdS}\). Then, we write the balanced equation for the reaction: \(\mathrm{Na}_{2}\mathrm{S} + \mathrm{CdSO}_{4} \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{4} + \mathrm{CdS ↓}\).

Key concepts

Solubility Rules

Solubility rules are essential guidelines to predict whether a substance will dissolve in a solvent like water. When two ionic compounds are mixed in solution, these rules help us determine if any insoluble compound will form, known as a precipitate. Here are some basic solubility rules:

• Most alkali metal salts (like those containing Na⁺ or K⁺) and ammonium salts (NH₄⁺) are soluble.
• All nitrate (NO₃^-) and acetate (CH₃COO^-) salts are soluble.
• Chlorides (Cl^-), bromides (Br^-), and iodides (I^-) are generally soluble, except with lead (Pb²⁺), silver (Ag⁺), and mercury (Hg₂²⁺).
• Sulfates (SO₄^2-) are usually soluble, except for those of barium (Ba²⁺), calcium (Ca²⁺), and lead (Pb²⁺).
• Hydroxides (OH^-) and sulfides (S^2-) are generally insoluble, except when paired with alkali metals or ammonium.

By applying these rules, we can often predict the formation of a precipitate in a chemical reaction. In our examples, we see that Cu(OH)₂ and CdS are the precipitates based on insolubility.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show the reactants transforming into the products. Each side of the equation includes chemical formulas, and the elements are represented by their symbols from the periodic table. A chemical equation should reflect what happens in a real chemical reaction in terms of molecules or moles.

• Reactants are substances that start a reaction.
• Products are substances formed from a reaction.

Consider the reaction between aqueous ext{NaCH}_{3} ext{COO} and ext{HCl}. The equation looks like this: a( ext{NaCH}_{3} ext{COO}) + b( ext{HCl}) ightarrow ext{products}. Each compound in the equation follows a specific pattern dictated by the reaction mechanism. Writing a chemical equation requires knowledge of the elemental composition of each reactant and product.

Ionic Compounds

Ionic compounds are formed through the electrostatic attraction between cations and anions. These compounds are usually formed between metals and nonmetals. For example, ext{NaCl} is an ionic compound formed by the attraction between ext{Na}^{+} and ext{Cl}^{-} ions. In a solution, ionic compounds often dissociate into individual ions. When solutions of different ionic compounds are mixed, the ions can rearrange. Sometimes an insoluble ionic compound forms, precipitating out of the solution. This precipitate is what we observe as a solid formation in the reaction. For instance, in the reaction between ext{KOH} and ext{Cu(NO}_{3})_{2}, the insoluble ext{Cu(OH)}_{2} is formed as a precipitate. This understanding of ionic interactions is crucial when predicting the products of chemical reactions.

Balanced Chemical Equations

Balanced chemical equations are crucial for accurately representing chemical reactions. To balance an equation, the number of atoms of each element on the reactant side must equal the number on the product side. This conservation reflects the Law of Conservation of Mass, stating that mass is neither created nor destroyed in a chemical reaction. For example, in the reaction ext{Na}_{2} ext{S} + ext{CdSO}_{4} ightarrow ext{Na}_{2} ext{SO}_{4} + ext{CdS}, both sides of the equation contain

• 2 Na atoms.
• 1 S atom.
• 1 Cd atom.
• 4 O atoms (from the sulfate ion).

The coefficients in front of the chemical formulas are adjusted to achieve balance, ensuring that the equation represents the reaction accurately. A correctly balanced equation provides the right proportions of reactants and products, which is essential for quantitative chemical calculations.