Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 23

Lead azide, \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2}\), is used as a detonator in car airbags. The impact of a collision causes \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2}\) to be converted into an enormous amount of gas that fills the airbag. At \(25^{\circ} \mathrm{C}\), a saturated solution of lead azide is prepared by dissolving \(25 \mathrm{mg}\) in water to make \(100.0 \mathrm{~mL}\) of solution. What is \(K_{\mathrm{sp}}\) for lead azide?

Short Answer

Expert verified
Answer: The solubility product constant (Ksp) for lead azide is 1.722 × 10^-9.

Step by step solution

01

Convert the mass of lead azide to moles.

To calculate the moles of lead azide dissolved, we will divide the given mass by the molar mass of lead azide. The molar mass of Pb(N3)2 is 331.21 g/mol (207.2 g/mol for Pb and 2 * 42.01 g/mol for N3). Thus: moles of lead azide = (25 mg) * (1 g / 1000 mg) * (1 mol / 331.21 g) moles of lead azide = 7.547 × 10^-5 mol
02

Find the molarity of lead azide in the solution.

To calculate the molarity of lead azide, we will divide the moles of lead azide by the volume of the solution in liters. Molarity = (moles of lead azide) / (volume of solution in L) Molarity = (7.547 × 10^-5 mol) / (0.100 L) Molarity = 7.547 × 10^-4 mol/L
03

Write the balanced equation for the dissolution of lead azide.

The dissolution of lead azide in water can be written as: Pb(N3)2 (s) <=> Pb^2+ (aq) + 2N3^- (aq)
04

Write the expression for Ksp.

The Ksp expression for the dissolution of lead azide is: Ksp = [Pb²⁺][N₃⁻]²
05

Calculate Ksp using the molar concentrations of ions found in the solution.

Since one mole of lead azide produces one mole of Pb²⁺ and two moles of N₃⁻, the molar concentrations will be: [Pb²⁺] = 7.547 × 10^-4 mol/L [N₃⁻] = 2 * 7.547 × 10^-4 mol/L = 1.509 × 10^-3 mol/L Now we can plug these values into the Ksp expression: Ksp = (7.547 × 10^-4)(1.509 × 10^-3)² Ksp = 1.722 × 10^-9 Thus, the solubility product constant (Ksp) for lead azide is 1.722 × 10^-9.

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.

Start your free trial

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

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

Before lead in paint was discontinued, lead chromate was a common pigment in yellow paint. A \(1.0\) -L solution is prepared by mixing \(0.50 \mathrm{mg}\) of lead nitrate with \(0.020 \mathrm{mg}\) of potassium chromate. Will a precipitate form? What should \(\left[\mathrm{Pb}^{2+}\right]\) be to just start precipitation?

Calculate the molar solubility of gold(I) chloride \(\left(K_{s p}=2.0 \times 10^{-13}\right)\) in \(0.10 \mathrm{M} \mathrm{NaCN}\). The complex ion formed is \(\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}\) with \(K_{\mathrm{f}}=2 \times 10^{38}\). Ignore any other competing equilibrium systems.

Consider the insoluble salts \(\mathrm{JQ} \mathrm{K}_{2} \mathrm{R}, \mathrm{L}_{2} \mathrm{~S}_{3}, \mathrm{MT}_{2}\), and \(\mathrm{NU}_{3}\). They are formed from the metal ions \(\mathrm{J}^{+}, \mathrm{K}^{+}, \mathrm{L}^{3+}, \mathrm{M}^{2+}\), and \(\mathrm{N}^{3+}\) and the nonmetal ions \(\mathrm{Q}^{-}, \mathrm{R}^{2-}, \mathrm{S}^{2-}, \mathrm{T}^{-}\), and \(\mathrm{U}^{-}\). All the salts have the same \(K_{\text {sp }}, 1 \times 10^{-10}\), at \(25^{\circ} \mathrm{C}\). (a) Which salt has the highest molar solubility? (b) Does the salt with the highest molar solubility have the highest solubility in g salt/100 g water? (c) Can the solubility of each salt in \(\mathrm{g} / 100 \mathrm{~g}\) water be determined from the information given? If yes, calculate the solubility of each salt in \(\mathrm{g} / 100 \mathrm{~g}\) water. If no, why not?

Consider the following solubility data for calcium oxalate \(\left(\mathrm{Ca} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) : $$ \begin{aligned} &K_{s p} \text { at } 25^{\circ} \mathrm{C}=4 \times 10^{-9} \\ &K_{s p} \text { at } 95^{\circ} \mathrm{C}=1 \times 10^{-8} \end{aligned} $$ Five hundred mL of a saturated solution is prepared at \(95^{\circ} \mathrm{C}\). How many milligrams of \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) will precipitate when the solution is cooled to \(25^{\circ} \mathrm{C}\) ? (Assume that supersaturation does not take place.)

Which of the following statements are true? (a) For an insoluble metallic salt, \(K_{\text {sp }}\) is always less than 1 . (b) More \(\mathrm{PbCl}_{2}\) can be dissolved at \(100^{\circ} \mathrm{C}\) than at \(25^{\circ} \mathrm{C}\). One can conclude that dissolving \(\mathrm{PbCl}_{2}\) is an exothermic process. (c) When strips of copper metal are added to a saturated solution of \(\mathrm{Cu}(\mathrm{OH})_{2}\), a precipitate of \(\mathrm{Cu}(\mathrm{OH})_{2}\) can be expected to form because of the common ion effect.
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation
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