/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 A 1.00-L, solution saturated at ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 56

A 1.00-L, solution saturated at \(25^{\circ} \mathrm{C}\) with lead(II) iodide contains \(0.54 \mathrm{~g}\) of \(\mathrm{Pbl}_{2}\). Calculate the solubility- product constant for this salt at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The solubility product constant (Ksp) for lead(II) iodide (PbI2) at 25掳C is approximately \(6.43 \times 10^{-9}\).

Step by step solution

01

Calculate the molar concentration of PbI2

We are given that a 1.00 L saturated solution contains 0.54 g of PbI2. To find the molar concentration, we will first convert the mass of PbI2 to moles and then divide by the volume of the solution in liters. Molecular weight of PbI2 = 207.2 (Pb) + 2 * 126.9 (I) = 460.0 g/mol Number of moles of PbI2 = \(\frac{0.54 \mathrm{~g}}{460.0 \mathrm{~g/mol}}\) = \(1.17 \times 10^{-3} \mathrm{mol}\) Molar concentration of PbI2 = \(\frac{1.17 \times 10^{-3} \mathrm{mol}}{1.00 \mathrm{~L}}\) = \(1.17 \times 10^{-3} \mathrm{M}\)
02

Write the balanced chemical equation and solubility product expression

For lead(II) iodide, the balanced chemical equation is: \[ \mathrm{PbI_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2I^{-} (aq)}\] The corresponding solubility product expression is: \[ K_{sp} = [\mathrm{Pb}^{2+}] [\mathrm{I}^{-}]^2\]
03

Determine the concentrations of Pb虏鈦� and I鈦� ions

Since the molar concentration of PbI2 is \(1.17 \times 10^{-3} \mathrm{M}\), the concentrations of the dissociated ions are as follows: - 1 mole of PbI2 gives 1 mole of Pb虏鈦� ions: \([\mathrm{Pb}^{2+}] = 1.17 \times 10^{-3} \mathrm{M}\) - 1 mole of PbI2 gives 2 moles of I鈦� ions: \([\mathrm{I}^{-}] = 2 \times 1.17 \times 10^{-3} \mathrm{M} = 2.34 \times 10^{-3} \mathrm{M}\)
04

Calculate the solubility product constant (Ksp)

Now, we will use the solubility product expression with the calculated ion concentrations to find Ksp: \[ K_{sp} = [\mathrm{Pb}^{2+}] [\mathrm{I}^{-}]^2 = (1.17 \times 10^{-3}) (2.34 \times 10^{-3})^2 = 6.43 \times 10^{-9} \] So, the solubility product constant for lead(II) iodide at 25掳C is approximately \(6.43 \times 10^{-9}\).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Lead(II) Iodide Solubility
The solubility of lead(II) iodide (PbI鈧�) refers to the amount of this compound that can dissolve in a solvent, creating a saturated solution. In a saturated solution, the maximum amount of solute is dissolved at a given temperature and pressure.
Lead(II) iodide is slightly soluble in water, meaning only a small amount dissolves before equilibrium is reached between the dissolved ions and the undissolved solid.
  • The dissolving process involves the dissociation of PbI鈧� into its constituent ions, piece by piece, in the water.
  • Because PbI鈧� is composed of lead (Pb虏鈦�) and iodide (I鈦�) ions, it dissolves by releasing these ions into the solution.
The solubility of a substance like lead(II) iodide can change with the temperature. Typically, increasing the temperature allows more PbI鈧� to dissolve, as more energy is available to break the ionic bonds. Understanding solubility is crucial since it underpins many reactions and processes in chemistry.
Molar Concentration Calculation
Calculating molar concentration helps to quantify the amount of a solute (like PbI鈧�) in a given volume of solution. The solution's concentration indicates how much solute is present, which is key for predicting the behavior of solutes in reactions and solutions.
In the problem, we start by converting the mass of PbI鈧� to moles since molarity is defined in terms of moles per liter.
  • First, you calculate the molar mass of PbI鈧� by adding the atomic masses of lead and iodide from the periodic table: 207.2 g/mol for lead and 126.9 g/mol for each iodide ion.

  • The total molar mass is 460.0 g/mol for PbI鈧�.

  • Next, divide the given mass by the molar mass: \( rac{0.54 ext{ g}}{460.0 ext{ g/mol}} = 1.17 imes 10^{-3} ext{ mol} \).

  • Finally, to find the molar concentration or molarity, divide the moles by the volume of the solution (in liters): \( rac{1.17 imes 10^{-3} ext{ mol}}{1.00 ext{ L}} = 1.17 imes 10^{-3} ext{ M} \).
Knowing the molar concentration is instrumental for determining the solubility product, helping to predict the solubility and behavior of ions in equilibrium.
Solubility Equilibrium
Solubility equilibrium is a dynamic state established when the rate of PbI鈧� dissolving equals the rate of Pb虏鈦� and I鈦� ions recombining into the solid. This balance allows us to define the solubility product constant, or Ksp.
In the case of lead(II) iodide, the equilibrium can be represented by the equation: \(\mathrm{PbI_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2I^{-} (aq)}\).
  • Ksp is a way to express the concentrations of the ions in a saturated solution at equilibrium directly through their concentrations.

  • For PbI鈧�, the expression is \( K_{sp} = [Pb^{2+}] [I^{-}]^2 \), showing how each ion contributes to the equilibrium constant.

  • The equilibrium expression indicates that a change in the concentration of either ion will impact the system until a new equilibrium is reached, maintaining the value of Ksp consistent under constant conditions.
By calculating Ksp, we can derive meaningful insights into the extent to which PbI鈧� can dissolve, a fundamental part of solubility and precipitation reactions.

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

A sample of \(0.2140 \mathrm{~g}\) of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\). of water and titrated with \(0.0950 \mathrm{M}\) \(\mathrm{NaOH}\). The acid required \(27.4 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(15.0 \mathrm{~mL}\) of base had been added in the titration, the \(\mathrm{pH}\) was found to be 6.50. What is the \(K_{a}\) for the unknown acid?

Consider the titration of \(30.0 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{NH}\) with \(0.025 \mathrm{M}\) \(\mathrm{HCl}\). Calculate the \(\mathrm{pH}\) after the following volumes of titrant have been added: (a) \(0 \mathrm{~mL}\), (b) \(20.0 \mathrm{~mL}\), (c) \(59.0 \mathrm{~mL}\), (d) \(60.0 \mathrm{~mL}\) (e) \(61.0 \mathrm{~mL}\) (f) \(65.0 \mathrm{~mL}\)

You are asked to prepare a pH \(=4.00\) buffer starting from \(1.50 \mathrm{~L}\) of \(0.0200 \mathrm{M}\) solution of benzoic acid \(\left(\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{COOH}\right)\) and any amount you need of sodium benzoate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COONa}\right)\). (a) What is the \(\mathrm{pH}\) of the benzoic acid solution prior to adding sodium benzoate? (b) How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.

Baking soda (sodium bicarbonate, \(\mathrm{NaHCO}_{3}\) ) reacts with acids in foods to form carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\), which in turn decomposes to water and carbon dioxide gas. In a cake batter, the \(\mathrm{CO}_{2}(\mathrm{~g})\) forms bubbles and causes the cake to rise. (a) A rule of thumb in baking is that \(1 / 2\) teaspoon of baking soda is neutralized by one cup of sour milk. The acid component in sour milk is lactic acid, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\). Write the chemical equation for this neutralization reaction. (b) The density of baking soda is \(2.16 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate the concentration of lactic acid in one cup of sour milk (assuming the rule of thumb applies), in units of \(\mathrm{mol} / \mathrm{L}\). (One cup \(=236.6 \mathrm{~mL}=48\) teaspoons). (c) If 1/2 teaspoon of baking soda is indeed completely neutralized by the lactic acid in sour milk, calculate the volume of carbon dioxide gas that would be produced at 1 atm pressure, in an oven set to \(350^{\circ} \mathrm{F}\).

A 1.00- \(\mathrm{L}\) solution saturated at \(25^{\circ} \mathrm{C}\) with calcium oxalate \(\left(\mathrm{CaC}_{2} \mathrm{O}_{4}\right)\) contains \(0.0061 \mathrm{~g}\) of \(\mathrm{CaC}_{2} \mathrm{O}_{4}\). Calculate the solubility-product constant for this salt at \(25^{\circ} \mathrm{C}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

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.