Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN: 9781305079373
Author: William L. Masterton, Cecile N. Hurley
Publisher: Cengage Learning
Question
Book Icon
Chapter 18, Problem 81QAP
Interpretation Introduction

Interpretation:

The number of years should be calculated which takes a 1.00 M solution of U to reach pH of 6.0.

Concept introduction:

Activity is defined as the product of rate constant and number of atoms. It is also defined as the rate of decay or expresses number of atoms decays in unit time.

The mathematical expression is given by:

A = kN

Where, A = activity, k = first order rate constant and N = number of radioactive nuclei.

Activity can be expressed in following units:

1 Bq = 1 atom/s

1 Ci = 3.700 × 1010 atoms/s

To express the small number, p scale is used, which implies to take the log of a number. Since, the concentration of [H+] in aqueous solution is small, by using the p scale in the form of pH scale, it is better way to represent acidity of solution.

Thus, pH = -log [H+]

The relation between half-life and decay constant is given by:

λ=0.693t1/2

Where, λ = decay constant and t1/2 = half-life

Expert Solution & Answer
Check Mark

Answer to Problem 81QAP

The 1.00 M solution of U will take t=3.03×106 year to achieve pH = 6.0.

Explanation of Solution

Given information:

Half-life of uranium = 7×108 years

The given equation is:

Zn(H2O)42+(aq)H+(aq)+Zn(H2O)3(OH)+(aq)      Ka=3.3×1010

The reaction between zinc ions and water is given as:

Zn(H2O)42+(aq)H+(aq)+Zn(H2O)3(OH)+(aq) 

Given pH value is 6.0.

pH = -log [H+]

Put the value of pH in above formula.

6.0 = -log [H+]

[H+] = 106.0

[H+] = 1×106

Now, let x be the concentration of Zn(H2O)42+.

Equilibrium constant (Ka)=3.3×1010

The formula of equilibrium constant is given by:

Ka=[H+][Zn(H2O)3(OH)+][Zn(H2O)42+]

Put the values in above formula.

3.3×1010=(1×106)(1×106)x1×106

Neglect, 1×106 in denominator

3.3×1010=(1×106)(1×106)x

3.3×1010x=(1×106)(1×106)

3.3×1010x=1×1012

x=1×10123.3×1010=0.003

Final concentration of U = 1 M0.003 M=0.997 M

The relation between half-life and decay constant is given by:

λ=0.693t1/2

λ=0.6937×108 years=9.9×1010year1

The formula for calculating time is given by:

Nt=No×eλt

Where, Nt = number of moles of radioactive element at t time (For uranium = 0.997)

No = number of moles of radioactive element at initial time (For uranium = 1.00)

Now, rearrange the formula in terms of time (t)

t=1λln(NoNt)

Put the values.

t=19.9×1010 year1ln(1.000.997)

t=19.9×1010 year1ln(1.000.997)

t=3.03×106 year

Thus, the 1.00 M solution of U will take t=3.03×106 year to achieve pH = 6.0.

Conclusion

The 1.00 M solution of U will take t=3.03×106 year to achieve pH = 6.0.

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
(a 4 shows scanning electron microscope (SEM) images of extruded actions of packing bed for two capillary columns of different diameters, al 750 (bottom image) and b) 30-μm-i.d. Both columns are packed with the same stationary phase, spherical particles with 1-um diameter. A) When the columns were prepared, the figure shows that the column with the larger diameter has more packing irregularities. Explain this observation. B) Predict what affect this should have on band broadening and discuss your prediction using the van Deemter terms. C) Does this figure support your explanations in application question 33? Explain why or why not and make any changes in your answers in light of this figure. Figure 4 SEM images of sections of packed columns for a) 750 and b) 30-um-i.d. capillary columns.³
fcrip = ↓ bandwidth Il temp 32. What impact (increase, decrease, or no change) does each of the following conditions have on the individual components of the van Deemter equation and consequently, band broadening? Increase temperature Longer column Using a gas mobile phase instead of liquid Smaller particle stationary phase Multiple Paths Diffusion Mass Transfer
34. Figure 3 shows Van Deemter plots for a solute molecule using different column inner diameters (i.d.). A) Predict whether decreasing the column inner diameters increase or decrease bandwidth. B) Predict which van Deemter equation coefficient (A, B, or C) has the greatest effect on increasing or decreasing bandwidth as a function of i.d. and justify your answer. Figure 3 Van Deemter plots for hydroquinone using different column inner diameters (i.d. in μm). The data was obtained from liquid chromatography experiments using fused-silica capillary columns packed with 1.0-μm particles. 35 20 H(um) 큰 20 15 90 0+ 1500 100 75 550 01 02 594 05 μ(cm/sec) 30 15 10

Chapter 18 Solutions

Chemistry: Principles and Reactions

Ch. 18 - Prob. 11QAPCh. 18 - Prob. 12QAPCh. 18 - Prob. 13QAPCh. 18 - Prob. 14QAPCh. 18 - Prob. 15QAPCh. 18 - Prob. 16QAPCh. 18 - Prob. 17QAPCh. 18 - Prob. 18QAPCh. 18 - Balance the following equations by filling in the...Ch. 18 - Prob. 20QAPCh. 18 - Prob. 21QAPCh. 18 - Prob. 22QAPCh. 18 - Prob. 23QAPCh. 18 - Prob. 24QAPCh. 18 - Prob. 25QAPCh. 18 - Prob. 26QAPCh. 18 - Prob. 27QAPCh. 18 - Prob. 28QAPCh. 18 - Prob. 29QAPCh. 18 - Prob. 30QAPCh. 18 - Prob. 31QAPCh. 18 - Prob. 32QAPCh. 18 - Prob. 33QAPCh. 18 - Prob. 34QAPCh. 18 - Prob. 35QAPCh. 18 - Prob. 36QAPCh. 18 - Prob. 37QAPCh. 18 - Prob. 38QAPCh. 18 - Prob. 39QAPCh. 18 - Prob. 40QAPCh. 18 - Prob. 41QAPCh. 18 - Prob. 42QAPCh. 18 - Prob. 43QAPCh. 18 - Prob. 44QAPCh. 18 - Prob. 45QAPCh. 18 - Prob. 46QAPCh. 18 - Prob. 47QAPCh. 18 - Prob. 48QAPCh. 18 - Prob. 49QAPCh. 18 - Prob. 50QAPCh. 18 - Prob. 51QAPCh. 18 - Prob. 52QAPCh. 18 - Prob. 53QAPCh. 18 - Prob. 54QAPCh. 18 - Prob. 55QAPCh. 18 - Prob. 56QAPCh. 18 - Prob. 57QAPCh. 18 - Prob. 58QAPCh. 18 - Prob. 59QAPCh. 18 - Prob. 60QAPCh. 18 - Prob. 61QAPCh. 18 - Prob. 62QAPCh. 18 - Prob. 63QAPCh. 18 - Prob. 64QAPCh. 18 - Prob. 65QAPCh. 18 - Prob. 66QAPCh. 18 - Prob. 67QAPCh. 18 - Prob. 68QAPCh. 18 - Prob. 69QAPCh. 18 - Prob. 70QAPCh. 18 - Prob. 71QAPCh. 18 - Prob. 72QAPCh. 18 - Fill in the following table:Ch. 18 - Prob. 74QAPCh. 18 - Prob. 75QAPCh. 18 - Prob. 76QAPCh. 18 - Prob. 77QAPCh. 18 - Prob. 78QAPCh. 18 - Prob. 79QAPCh. 18 - Carbon-14 (C-14) with a half-life of 5730 years...Ch. 18 - Prob. 81QAPCh. 18 - Prob. 82QAP
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Introductory Chemistry: A Foundation
Chemistry
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
Chemistry: Matter and Change
Chemistry
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Glencoe/McGraw-Hill School Pub Co