EBK CHEMICAL PRINCIPLES
EBK CHEMICAL PRINCIPLES
8th Edition
ISBN: 9781305856745
Author: DECOSTE
Publisher: CENGAGE LEARNING - CONSIGNMENT
bartleby

Videos

Question
Book Icon
Chapter 17, Problem 133CP
Interpretation Introduction

Interpretation: The expected osmotic pressure of the given solution needs to be calculated.

Concept introduction: The excess pressure applied to a solution in order to prevent the passage of solvent to the solution is said to be osmotic pressure. The formula to determine the osmotic pressure, π is:

  π = iMRT

Where i is van’t Hoff’s factor, M is molar concentration, R is ideal gas constant and T is temperature in K.

Expert Solution
Check Mark

Answer to Problem 133CP

The expected osmotic pressure of the given solution is 6.11 atm.

Explanation of Solution

Given:

0.05 mole of Fe2(SO4)3 is dissolved in water to make a solution of 1.0 L. The ions present in solution is SO42- (aq) and Fe(H2O)63+(aq) ions.

The Fe(H2O)63+ behaves as an acid as:

  Fe(H2O)63+   Fe(OH)(H2O)52+ + H+

Converting temperature from degree Celsius to Kelvin as

  25+273 = 298 K

The dissociation reaction of Fe2(SO4)3 is:

  Fe2(SO4)3(aq)  2Fe3+(aq) + 3SO42(aq)

From the above reaction it is observed that the number of ions formed are 5. So, i = 5 .

Now, the osmotic pressure is calculated using formula:

  π = iMRT

Substituting the values:

  π = 5(0.05 mol/L)(0.08206 LatmK1mol1)(298 K)π = 6.11 atm

Hence, the expected osmotic pressure of the given solution is 6.11 atm.

Interpretation Introduction

Interpretation: The Ka for the dissociation reaction of Fe(H2O)63+ needs to be calculated.

Concept introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

  K = concentration of productsconcentration of reactants

Expert Solution
Check Mark

Answer to Problem 133CP

The Ka for the dissociation reaction of Fe(H2O)63+ is 8.3×103 .

Explanation of Solution

The contribution of each ion in the osmotic pressure of the solution of Fe2(SO4)3 is determined as follows:

The dissociation reaction of Fe2(SO4)3 is:

  Fe2(SO4)3(aq)  2Fe3+(aq) + 3SO42(aq)

From the above reaction the contribution of each ion in the osmotic pressure is calculated as:

  Fe3+ = 25 × 6.11 atmFe3+ = 2.44 atm

  SO42 = 35 × 6.11 atmSO42 = 3.67 atm

Since, the actual osmotic pressure is 6.73 and the ion that remain same in both the dissociation is SO42 so, the contribution of SO42 ions in total osmotic pressure is the same.

The contribution of Fe(H2O)63+ = 6.73 - 3.67 = 3.06 atm .

The initial concentration of Fe(H2O)63+ is:

  Fe(H2O)63+ = 2(0.0500 M) = 0.1000 M

Creating ICE table for weak acid, Fe(H2O)63+ as:

                                Fe(H2O)63+                         H+          +        Fe(OH)(H2O)52+Initial:                     0.100 M                                 0                               0Change:                   -x                                            +x                            +xEquilibrium:           0.100 - x                                 x                               x

Now, calculating the total ion concentration as:

  πRT= 0.100 M - x + x + xπRT = 0.100 M + x

Substituting the values:

  x = 3.06 atm0.08206 Latmmol1K1(298 K) - 0.100 Mx = 0.025 M

Now, the value of acid dissociation constant is calculated using formula:

  K = concentration of productsconcentration of reactants

So, the expression for acid dissociation constant for the above reaction is:

  Ka = [H+][Fe(OH)(H2O)52+][Fe(H2O)63+]

Substituting the values:

  Ka = x20.100 - xKa = (0.025)2(0.100 - 0.025)Ka = 0.0006250.075Ka = 8.3×103

Thus, the Ka for the dissociation reaction of Fe(H2O)63+ is 8.3×103 .

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!
Students have asked these similar questions
Arrange the solutions in order of increasing acidity. (Note that K (HF) = 6.8 x 10 and K (NH3) = 1.8 × 10-5) Rank solutions from least acidity to greatest acidity. To rank items as equivalent, overlap them. ▸ View Available Hint(s) Least acidity NH&F NaBr NaOH NH,Br NaCIO Reset Greatest acidity
1. Consider the following molecular-level diagrams of a titration. O-HA molecule -Aion °° о ° (a) о (b) (c) (d) a. Which diagram best illustrates the microscopic representation for the EQUIVALENCE POINT in a titration of a weak acid (HA) with sodium. hydroxide? (e)
Answers to the remaining 6 questions will be hand-drawn on paper and submitted as a single file upload below: Review of this week's reaction: H₂NCN (cyanamide) + CH3NHCH2COOH (sarcosine) + NaCl, NH4OH, H₂O ---> H₂NC(=NH)N(CH3)CH2COOH (creatine) Q7. Draw by hand the reaction of creatine synthesis listed above using line structures without showing the Cs and some of the Hs, but include the lone pairs of electrons wherever they apply. (4 pts) Q8. Considering the Zwitterion form of an amino acid, draw the Zwitterion form of Creatine. (2 pts) Q9. Explain with drawing why the C-N bond shown in creatine structure below can or cannot rotate. (3 pts) NH2(C=NH)-N(CH)CH2COOH This bond Q10. Draw two tautomers of creatine using line structures. (Note: this question is valid because problem Q9 is valid). (4 pts) Q11. Mechanism. After seeing and understanding the mechanism of creatine synthesis, students should be ready to understand the first half of one of the Grignard reactions presented in a past…

Chapter 17 Solutions

EBK CHEMICAL PRINCIPLES

Ch. 17 - Prob. 11DQCh. 17 - Prob. 12ECh. 17 - Prob. 13ECh. 17 - Prob. 14ECh. 17 - Prob. 15ECh. 17 - Prob. 16ECh. 17 - Prob. 17ECh. 17 - Prob. 18ECh. 17 - Prob. 19ECh. 17 - Prob. 20ECh. 17 - Prob. 21ECh. 17 - Prob. 22ECh. 17 - Prob. 23ECh. 17 - Prob. 24ECh. 17 - Prob. 25ECh. 17 - Prob. 26ECh. 17 - Prob. 27ECh. 17 - Prob. 28ECh. 17 - Prob. 29ECh. 17 - Prob. 30ECh. 17 - Prob. 31ECh. 17 - Prob. 32ECh. 17 - Prob. 33ECh. 17 - Prob. 34ECh. 17 - Prob. 35ECh. 17 - Prob. 36ECh. 17 - Prob. 37ECh. 17 - Prob. 38ECh. 17 - Prob. 39ECh. 17 - Prob. 40ECh. 17 - Rationalize the temperature dependence of the...Ch. 17 - Prob. 42ECh. 17 - Prob. 43ECh. 17 - Prob. 44ECh. 17 - Prob. 45ECh. 17 - Prob. 46ECh. 17 - Prob. 47ECh. 17 - Prob. 48ECh. 17 - Prob. 49ECh. 17 - Prob. 50ECh. 17 - Prob. 51ECh. 17 - Prob. 52ECh. 17 - Prob. 53ECh. 17 - Prob. 54ECh. 17 - Prob. 55ECh. 17 - Prob. 56ECh. 17 - The following plot shows the vapor pressure of...Ch. 17 - Prob. 58ECh. 17 - Prob. 59ECh. 17 - Prob. 60ECh. 17 - Prob. 61ECh. 17 - Prob. 62ECh. 17 - Prob. 63ECh. 17 - Prob. 64ECh. 17 - Prob. 65ECh. 17 - Prob. 66ECh. 17 - Prob. 67ECh. 17 - An aqueous solution of 10.00 g of catalase, an...Ch. 17 - Prob. 69ECh. 17 - What volume of ethylene glycol (C2H6O2) , a...Ch. 17 - Prob. 71ECh. 17 - Erythrocytes are red blood cells containing...Ch. 17 - Prob. 73ECh. 17 - Prob. 74ECh. 17 - Prob. 75ECh. 17 - Prob. 76ECh. 17 - Prob. 77ECh. 17 - Prob. 78ECh. 17 - Prob. 79ECh. 17 - Prob. 80ECh. 17 - Consider the following solutions: 0.010 m Na3PO4...Ch. 17 - From the following: pure water solution of...Ch. 17 - Prob. 83ECh. 17 - Prob. 84ECh. 17 - Prob. 85ECh. 17 - Prob. 86ECh. 17 - Prob. 87ECh. 17 - Prob. 88ECh. 17 - Prob. 89ECh. 17 - Prob. 90ECh. 17 - Prob. 91ECh. 17 - Prob. 92ECh. 17 - Prob. 93AECh. 17 - Prob. 94AECh. 17 - Prob. 95AECh. 17 - Prob. 96AECh. 17 - The term proof is defined as twice the percent by...Ch. 17 - Prob. 98AECh. 17 - Prob. 99AECh. 17 - Prob. 100AECh. 17 - Prob. 101AECh. 17 - Prob. 102AECh. 17 - Prob. 103AECh. 17 - Prob. 104AECh. 17 - Prob. 105AECh. 17 - Prob. 106AECh. 17 - Prob. 107AECh. 17 - Prob. 108AECh. 17 - Prob. 109AECh. 17 - Prob. 110AECh. 17 - Prob. 111AECh. 17 - Prob. 112AECh. 17 - Prob. 113AECh. 17 - Prob. 114AECh. 17 - Formic acid (HCO2H) is a monoprotic acid that...Ch. 17 - Prob. 116AECh. 17 - Prob. 117AECh. 17 - Prob. 118AECh. 17 - Prob. 119AECh. 17 - Prob. 120AECh. 17 - Prob. 121AECh. 17 - Prob. 122AECh. 17 - Prob. 123AECh. 17 - Prob. 124AECh. 17 - Prob. 125AECh. 17 - Prob. 126AECh. 17 - Prob. 127CPCh. 17 - Prob. 128CPCh. 17 - Prob. 129CPCh. 17 - Plants that thrive in salt water must have...Ch. 17 - Prob. 131CPCh. 17 - Prob. 132CPCh. 17 - Prob. 133CPCh. 17 - Prob. 134CPCh. 17 - Prob. 135CPCh. 17 - Prob. 136CP
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.
Similar questions
SEE MORE QUESTIONS
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: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781133611097
Author:Steven S. Zumdahl
Publisher:Cengage Learning
Text book image
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning
Text book image
General, Organic, and Biological Chemistry
Chemistry
ISBN:9781285853918
Author:H. Stephen Stoker
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Ocean Chemistry; Author: Beverly Owens;https://www.youtube.com/watch?v=IDQzklIr57Q;License: Standard YouTube License, CC-BY