Ch.4 - Reactions in Aqueous Solution

Problem 5

Chapter 4, Problem 5

You are presented with a white solid and told that due to careless labeling it is not clear if the substance is barium chloride, lead chloride, or zinc chloride. When you transfer the solid to a beaker and add water, the solid dissolves to give a clear solution. Next an Na2SO41aq2 solution is added and a white precipitate forms. What is the identity of the unknown white solid?

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Identify the possible compounds: barium chloride (BaCl_2), lead chloride (PbCl_2), and zinc chloride (ZnCl_2).

Consider the solubility of each compound in water: BaCl_2 and ZnCl_2 are soluble, while PbCl_2 is slightly soluble.

Since the solid dissolves completely in water, it is likely BaCl_2 or ZnCl_2.

Add Na_2SO_4 to the solution: BaSO_4 is insoluble and will form a white precipitate, while ZnSO_4 is soluble.

The formation of a white precipitate upon adding Na_2SO_4 indicates the presence of Ba^2+ ions, identifying the unknown solid as barium chloride (BaCl_2).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Solubility and Dissolution

Solubility refers to the ability of a substance to dissolve in a solvent, forming a solution. In this case, the white solid dissolves in water, indicating it is soluble. The solubility of ionic compounds varies; for example, barium chloride, lead chloride, and zinc chloride all have different solubility characteristics in water, which is crucial for identifying the unknown solid.

Precipitation Reactions

A precipitation reaction occurs when two soluble salts react in solution to form an insoluble solid, known as a precipitate. In this scenario, the addition of Na2SO4 results in a white precipitate, which suggests that one of the ions from the unknown solid reacts with sulfate ions to form an insoluble compound. Identifying the precipitate can help determine the identity of the original solid.

Ionic Compounds and Their Solubility Rules

Ionic compounds consist of positively and negatively charged ions. Their solubility in water is governed by specific rules; for instance, sulfates are generally soluble except for those of barium, lead, and calcium. Understanding these rules allows us to predict the behavior of the unknown solid when mixed with Na2SO4, leading to the identification of the solid based on the formation of a precipitate.

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Solubility Rules

Related Practice

Textbook Question

Aqueous solutions of three different substances, AX, AY, and AZ, are represented by the three accompanying diagrams. Identify each substance as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. (a)

(b)

(c)

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Textbook Question

Use the molecular representations shown here to classify each compound as a nonelectrolyte, a weak electrolyte, or a strong electrolyte (see Figure 4.6 for the element color scheme). (a)

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Textbook Question

The concept of chemical equilibrium is very important. Which one of the following statements is the most correct way to think about equilibrium? (a) If a system is at equilibrium, nothing is happening. (b) If a system is at equilibrium, the rate of the forward reaction is equal to the rate of the back reaction. (c) If a system is at equilibrium, the product concentration is changing over time.

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Textbook Question

Which of the following ions will always be a spectator ion in a precipitation reaction? (a) Cl- (b) NO3- (c) NH4+ (d) S2- (e) SO42-

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Open Question

The labels have fallen off three bottles containing powdered samples of metals; one contains zinc, one contains lead, and the other contains platinum. You have three solutions at your disposal: 1 M sodium nitrate, 1 M nitric acid, and 1 M nickel nitrate. How could you use these solutions to determine the identities of each metal powder?

Open Question

In each of the following pairs, indicate which has the higher concentration of Cl^- ion: (a) 0.10 M AlCl₃ solution or a 0.25 M LiCl solution (b) 150 mL of a 0.05 M MnCl₃ solution or 200 mL of 0.10 M KCl solution (c) a 2.8 M HCl solution or a solution made by dissolving 23.5 g of KCl in water to make 100 mL of solution.