General Chemistry
General Chemistry
4th Edition
ISBN: 9781891389603
Author: Donald A. McQuarrie, Peter A. Rock, Ethan B. Gallogly
Publisher: University Science Books
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 22, Problem 22.25P
Interpretation Introduction

Interpretation:

The solubility in grams per liter of CuBr(s) in 0.50MNH3 should be calculated.

Concept Introduction:

Solubility is defined as the maximum quantity of solute dissolved in a given amount of solvent to make a saturated solution at a particular temperature.

Molar solubility is the number of moles of a solute that can be dissolved in one liter of a solution. It is expressed as mol/L or M (molarity).

Consider a general reaction:

  MnXm(s)nMm+(aq)+mXn+(aq)

The relation between solubility product and molar solubility is as follows:

  Ksp=[Mm+]n[Xn-]m

Here

The solubility product of salt is Ksp.

The molar solubility of Mm+ ion is [Mm+]

The molar solubility of Xn- ion is [Xn-]

Expert Solution & Answer
Check Mark

Answer to Problem 22.25P

The solubility in grams per liter of CuBr(s) is given as 35gL1.

Explanation of Solution

The two applicable equations are given below:

  CuBr(s)Cu+(aq)+Br(aq)Ksp=6.3×109M2(1)Cu+(aq)+2NH3[Cu(NH3)]2+(aq)Kf=6.3×1010M2(2)

Addition of the both equation will give,

  CuBr(s)+2NH3[Cu(NH3)]2+(aq)+Br(aq)(3)Kc=Ksp.Kf=6.3×109M2×6.3×1010M2=396.9

Since, Kc>>Ksp, equation (1) can be neglected and (3) can be used to form the concentration table.

  CuBr(s)+2NH3[Cu(NH3)]2+(aq)+Br(aq)Initial0.50M0M0MChange2x+x+xEquilibrium0.50M2xxx

The expression for equilibrium constant is,

  Kc=x2(0.50M2x)2=396.9x(0.50M2x)=19.9x=0.244[Cu(NH3)]2+=x=0.244[NH3]=0.50M2x=0.50M(2×0.244)=0.012

The concentration of Cu+ can be calculated as follows:

  Kf=[Cu(NH3)]2+[Cu+][NH3]2=6.3×1010M2[Cu+]=[Cu(NH3)]2+6.3×1010M2[NH3]2=0.244M6.3×1010M2[0.012M]2[Cu+]=2.7×108M

The total solubility of CuBr(s) can be calculated as follows:

  s=[Cu+]+[Cu(NH3)]2+=2.7×108M+0.244Ms0.244M

Therefore, the solubility of CuBr(s) in grams per liter is given below:

  s=0.244mol.L1(143.45g1molCuBr)s=35gL1

The solubility in grams per liter of CuBr(s) is calculated to be 35gL1.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 22 Solutions

General Chemistry

Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
General Chemistry | Acids & Bases; Author: Ninja Nerd;https://www.youtube.com/watch?v=AOr_5tbgfQ0;License: Standard YouTube License, CC-BY