Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The labels have fallen off two bottles, one containing \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) and the other containing \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\). You have a bottle of dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\). How could you use it to test a portion of each solution to identify which solution is which? [Section 4.2]

Short Answer

Expert verified
To identify which solution is Mg(NO3)2 and which is Pb(NO3)2 using dilute H2SO4, take small samples from each unknown solution and add H2SO4 to each. Observe the reactions: - If a white precipitate forms, the solution is Pb(NO3)2 (due to the formation of PbSO4). - If the solution remains clear, the solution is Mg(NO3)2 (due to the formation of soluble MgSO4).

Step by step solution

01

Write down the chemical reactions

When Mg(NO3)2 or Pb(NO3)2 reacts with H2SO4, the following reactions take place: \[Mg(NO_3)_2 + H_2SO_4 \rightarrow MgSO_4 + 2HNO_3\] and \[Pb(NO_3)_2 + H_2SO_4 \rightarrow PbSO_4 + 2HNO_3\]
02

Determine the difference in the reaction products

The products of the reactions are MgSO4 and PbSO4. The key difference between MgSO4 and PbSO4 is their solubility in water. MgSO4 is soluble in water and forms a clear solution, while PbSO4 is not very soluble and forms a white precipitate.
03

Perform the test with dilute H2SO4

To identify which solution is which, follow these steps: 1. Take two small samples, one from each of the two unknown solutions. 2. Add a small amount of dilute H2SO4 to each sample. 3. Observe the reactions for any formation of precipitate.
04

Analyze the results

After adding dilute H2SO4: - If a white precipitate is formed, the solution is Pb(NO3)2 (due to the formation of PbSO4). - If the solution remains clear, the solution is Mg(NO3)2 (due to the formation of soluble MgSO4). This way, you can identify which solution contains Mg(NO3)2 and which contains Pb(NO3)2.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility
Solubility is a critical property of substances, especially when discussing chemical reactions involving solutions. It refers to a substance's ability to dissolve in a solvent, like water. Solubility is influenced by factors such as temperature, pressure, and the nature of both the solute and solvent.

In the context of the given exercise, understanding solubility helps differentiate between the two products formed during the reactions.
  • Magnesium sulfate ( MgSO_4 ) is highly soluble in water. As a result, when it forms during the reaction, it readily dissolves, leaving the solution clear.
  • Lead sulfate ( PbSO_4 ), on the other hand, has low solubility in water. This means it does not dissolve easily, leading to the formation of a precipitate.
The differing solubility of MgSO_4 and PbSO_4 is the key to identifying the unknown solutions.
Precipitate Formation
Precipitation occurs when a solution reacts to form a solid, known as a precipitate. This is a crucial phenomenon in chemistry because it often indicates the formation of a new compound.

In this exercise, Pb(NO_3)_2 reacts with H_2SO_4 to form PbSO_4 , which is insoluble in water. As a result:
  • PbSO_4 appears as a white solid precipitate in the reaction mixture.
  • This solid formation suggests that the reaction mixture contains lead.
Being able to visually detect a precipitate allows chemists to identify substances present in a solution. The absence or presence of a precipitate is a simple yet powerful tool in chemical analysis.
Solution Identification
Identifying solutions based on their chemical properties, such as solubility and precipitate formation, is a practical application of chemistry.

In the scenario given, using H_2SO_4 to distinguish between Mg(NO_3)_2 and Pb(NO_3)_2 depends on observing the reaction's outcome:
  • If a white precipitate forms, the solution was Pb(NO_3)_2 , due to the creation of PbSO_4 , which is insoluble.
  • If no precipitate forms and the solution remains clear, the solution was Mg(NO_3)_2 because MgSO_4 is soluble.
This method showcases how chemical properties can be employed for solution identification without advanced equipment, relying solely on simple tests and visual observations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write balanced molecular and net ionic equations for the reactions of (a) manganese with dilute sulfuric acid; (b) chromium with hydrobromic acid; (c) tin with hydrochloric acid; (d) aluminum with formic acid, \(\mathrm{HCOOH}\)

Methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), and hydrogen chloride, \(\mathrm{HCl}\), are both molecular substances, yet an aqueous solution of methanol does not conduct an electrical current, whereas a solution of \(\mathrm{HCl}\) does conduct. Account for this difference. [Section 4.1]

A solid sample of \(\mathrm{Zn}(\mathrm{OH})_{2}\) is added to \(0.350 \mathrm{~L}\) of \(0.500\) \(M\) aqueous HBr. The solution that remains is still acidic. It is then titrated with \(0.500 \mathrm{M} \mathrm{NaOH}\) solution, and \(\mathrm{it}\) takes \(88.5 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution to reach the equivalence point. What mass of \(\mathrm{Zn}(\mathrm{OH})_{2}\) was added to the HBr solution?

Formic acid, HCOOH, is a weak electrolyte. What solute particles are present in an aqueous solution of this compound? Write the chemical equation for the ionization of \(\mathrm{HCOOH}\).

A sample of solid \(\mathrm{Ca}(\mathrm{OH})_{2}\) is stirred in water at \(30^{\circ} \mathrm{C}\) until the solution contains as much dissolved \(\mathrm{Ca}(\mathrm{OH})_{2}\) as it can hold. A \(100-\mathrm{mL}\) sample of this solution is withdrawn and titrated with \(5.00 \times 10^{-2} \mathrm{M} \mathrm{HBr}\). It requires \(48.8 \mathrm{~mL}\) of the acid solution for neutralization. What is the molarity of the \(\mathrm{Ca}(\mathrm{OH})_{2}\) solution? What is the solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\) in water, at \(30^{\circ} \mathrm{C}\), in grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) per \(100 \mathrm{~mL}\) of solution?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free