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

Short answer

The precipitates formed in each reaction are: (a) None (b) \(\mathrm{Cu}(\mathrm{OH})_{2}\) (c) \(\mathrm{CdS}\) The balanced equations for each reaction are: (a) \(\mathrm{NaCH}_{3} \mathrm{COO} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{HCH}_{3} \mathrm{COOH}\) (b) \(\mathrm{2KOH} + \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow \mathrm{Cu}(\mathrm{OH})_{2} + \mathrm{2KNO}_{3}\) (c) \(\mathrm{Na}_{2} \mathrm{S} + \mathrm{CdSO}_{4} \rightarrow \mathrm{CdS} + \mathrm{2Na}_{2} \mathrm{SO}_{4}\)

Step-by-step solution

Write the chemical reaction between the given reactants

For each set of reactants, we will write the chemical reactions: (a) \(\mathrm{NaCH}_{3} \mathrm{COO} + \mathrm{HCl}\) (b) \(\mathrm{KOH} + \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) (c) \(\mathrm{Na}_{2} \mathrm{S} + \mathrm{CdSO}_{4}\)

Determine the possible precipitate formations using solubility rules

Using solubility rules, we will evaluate the products formed in each reaction and identify if any of them form a precipitate.

Write the balanced equation for each reaction

(a) \(\mathrm{NaCH}_{3} \mathrm{COO} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{HCH}_{3} \mathrm{COOH}\) No precipitate formation in this reaction since both products, \(\mathrm{NaCl}\) and \(\mathrm{HCH}_{3} \mathrm{COOH}\) are soluble. (b) \(\mathrm{2KOH} + \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow \mathrm{Cu}(\mathrm{OH})_{2} + \mathrm{2KNO}_{3}\) A precipitate forms in this reaction: \(\mathrm{Cu}(\mathrm{OH})_{2}\) (c) \(\mathrm{Na}_{2} \mathrm{S} + \mathrm{CdSO}_{4} \rightarrow \mathrm{CdS} + \mathrm{2Na}_{2} \mathrm{SO}_{4}\) A precipitate forms in this reaction: \(\mathrm{CdS}\)

Identify the precipitate for each reaction

(a) No precipitate formation (b) Precipitate: \(\mathrm{Cu}(\mathrm{OH})_{2}\) (c) Precipitate: \(\mathrm{CdS}\) In conclusion, the precipitates formed in each reaction are: (a) None (b) \(\mathrm{Cu}(\mathrm{OH})_{2}\) (c) \(\mathrm{CdS}\) The balanced equations for each reaction are: (a) \(\mathrm{NaCH}_{3} \mathrm{COO} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{HCH}_{3} \mathrm{COOH}\) (b) \(\mathrm{2KOH} + \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow \mathrm{Cu}(\mathrm{OH})_{2} + \mathrm{2KNO}_{3}\) (c) \(\mathrm{Na}_{2} \mathrm{S} + \mathrm{CdSO}_{4} \rightarrow \mathrm{CdS} + \mathrm{2Na}_{2} \mathrm{SO}_{4}\)

Key concepts

Solubility Rules

Understanding solubility rules is critical when predicting the outcomes of chemical precipitation reactions. These rules are a set of guidelines that help us determine whether an ionic compound is likely to dissolve in water, making a solution, or form a solid precipitate. A precipitate is a solid that emerges from a liquid solution. The ability of a substance to dissolve depends on its solubility. For example, salts containing nitrate (o3) or alkali metal cations like sodium (a) and potassium (o3) are generally soluble. On the other hand, compounds containing silver (o3) or lead (o3) ions often lead to precipitate formation as they are less soluble.

When we mix solutions, the ions can react to form new compounds. If any of these compounds has low solubility, as dictated by the solubility rules, it will precipitate out of the solution. In the textbook problem provided, the exercise challenges students to apply these rules to predict precipitate formations in reactions. For example, when sodium acetate (o3NaCH_3COOo3) reacts with hydrochloric acid (o3HClo3), no precipitate forms because all possible products are soluble in water.

Balanced Chemical Equations

The heart of a chemical reaction is represented by its balanced chemical equation. It shows the substances that react (reactants) and the substances that are produced (products), with the number of atoms for each element balanced on both sides of the equation. Students must understand how to balance equations to conform to the Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction.

The equation must have the same number of atoms of each element on both sides. This step is crucial before one can identify the precipitate in a reaction. In our exercise, balancing equations ensures students understand the stoichiometry of the reaction. For instance, when potassium hydroxide (o3KOHo3) reacts with copper(II) nitrate (o3Cu(NO_3)_2o3), the balanced equation shows that two moles of o3KOHo3 react with one mole of o3Cu(NO_3)_2o3 to produce one mole of the precipitate, copper(II) hydroxide (o3Cu(OH)_2o3), and two moles of potassium nitrate (o3KNO_3o3), which stays dissolved in the solution.

Precipitate Identification

Once we've used solubility rules to determine which products are likely to form a precipitate, and written a balanced equation, the next step is precipitate identification. This process involves looking at the reaction products and discerning which one, if any, has formed a solid. The formation of a precipitate can often be detected visually as cloudiness or a solid deposit in the reaction mixture.

For students to successfully identify a precipitate, they must recognize the physical changes that occur during the reaction. Noticing a precipitate's characteristic color can also aid identification, such as the deep blue of copper(II) hydroxide (o3Cu(OH)_2o3) or the yellow of cadmium sulfide (o3CdSo3). Through the exercises, we guide students in predicting and confirming which products precipitate. For example, when sodium sulfide (o3Na_2So3) is mixed with cadmium sulfate (o3CdSO_4o3), cadmium sulfide precipitates due to its low solubility, following the known solubility rules.