/*! 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 52 A 1.00-L solution saturated at \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 52

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

Short Answer

Expert verified
The solubility product constant (K鈧�) for lead(II) iodide at 25掳C is 6.40 x 10鈦烩伖.

Step by step solution

01

Calculate the molar solubility of PbI鈧�

To find the molar solubility, we first need to calculate the number of moles of PbI鈧� present in the 1.00-L solution. Given mass of PbI鈧� = 0.54 g Molar mass of PbI鈧� = (Pb=207.2 g/mol) + 2*(I=126.9 g/mol) = 207.2 + (2*126.9) = 460.99 g/mol Moles of PbI鈧� = \(\frac{0.54 g}{460.99 g/mol}\) = \(1.17 \times 10^{-3} mol\) Now, we have the number of moles of PbI鈧�. We can find the molar solubility of PbI鈧� in the solution by dividing the moles of PbI鈧� by the volume of the solution. Molar solubility of PbI鈧� = \(\frac{1.17 \times 10^{-3} mol}{1.00 L}\) = 1.17 x 10鈦宦� mol/L
02

Determine the Balance Dissociation Equation for PbI鈧�

The balanced dissociation equation for PbI鈧� can be written as: PbI鈧�(s) \(\rightleftharpoons\) Pb虏鈦�(aq) + 2I鈦�(aq) This balanced equation tells us that for every one mole of Pb虏鈦� in the solution, there will also be two moles of I鈦�.
03

Calculating Concentration of Ions

We know that the molar solubility of PbI鈧� is 1.17 x 10鈦宦� mol/L, which is equal to the concentration of Pb虏鈦� in the solution. Concentration of Pb虏鈦� = 1.17 x 10鈦宦� mol/L Since for every mole of Pb虏鈦� there are two moles of I鈦�, the concentration of I鈦� will be: Concentration of I鈦� = 2 x (1.17 x 10鈦宦�) = 2.34 x 10鈦宦� mol/L
04

Calculate the solubility product constant (K鈧�)

Now that we have the concentration of Pb虏鈦� and I鈦�, we can find the solubility product constant K鈧� using the following expression: K鈧� = [Pb虏鈦篯 脳 [I鈦籡虏 Plugging in the values, we get: K鈧� = (1.17 x 10鈦宦�) 脳 (2.34 x 10鈦宦�)虏 = 6.40 x 10鈦烩伖 Thus, the solubility product constant for lead(II) iodide at 25掳C is 6.40 x 10鈦烩伖.

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!

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

How many milliliters of \(0.105 \mathrm{M} \mathrm{HCl}\) are needed to titrate each of the following solutions to the equivalence point: (a) \(45.0 \mathrm{~mL}\) of \(0.0950 \mathrm{M} \mathrm{NaOH}\), (b) \(22.5 \mathrm{~mL}\) of \(0.118 \mathrm{M} \mathrm{NH}_{3}\), (c) \(125.0 \mathrm{~mL}\) of a solution that contains \(1.35 \mathrm{~g}\) of \(\mathrm{NaOH}\) per liter?

(a) What is the ratio of \(\mathrm{HCO}_{3}^{-}\) to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in blood of pH \(7.4\) ? (b) What is the ratio of \(\mathrm{HCO}_{3}^{-}\) to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in an exhausted marathon runner whose blood pH is 7.1?

Fluoridation of drinking water is employed in many places to aid in the prevention of dental caries. Typically the \(\mathrm{F}^{-}\) ion concentration is adjusted to about \(1 \mathrm{ppb}\). Some water supplies are also "hard"; that is, they contain certain cations such as \(\mathrm{Ca}^{2+}\) that interfere with the action of soap. Consider a case where the concentration of \(\mathrm{Ca}^{2+}\) is 8 ppb. Could a precipitate of \(\mathrm{CaF}_{2}\) form under these conditions? (Make any necessary approximations.)

Derive an equation similar to the HendersonHasselbalch equation relating the pOH of a buffer to the \(\mathrm{p} K_{b}\) of its base component.

If \(40.00 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) is titrated with \(0.100 \mathrm{M}\) \(\mathrm{HCl}\), calculate (a) the \(\mathrm{pH}\) at the start of the titration; (b) the volume of \(\mathrm{HCl}\) required to reach the first equivalence point and the predominant species present at this point; (c) the volume of \(\mathrm{HCl}\) required to reach the second equivalence point and the predominant species present at this point; (d) the pH at the second equivalence point.
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

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.