Assigned: 1-Mar-17
1st Due Date: 8-Mar-17
2nd Due Date: 10-Mar-17
Lecture 11: Speciation
Lecture 12: Uranium chemistry
Lecture 13: Neptunium chemistry
Lecture 14: Plutonium chemistry
Lecture 15: Americium and Curium chemistry
Lecture 16: Chemistry in reactor fuel
Lecture 16: Chemistry in reactor fuel
Lecture 17: Separations
Lecture 18: Application of Nuclear Material
Lecture 19: Nuclear Forensics
Use lecture notes, textbooks, Chart of the Nuclides, Table of the Isotopes, and web pages. Use the chart of the nuclides as your primary dataset for isotope half-life. Show your work or references on a separate page and save electronically. Submission of the work is not required for the 1st due date. Please use 3 significant digits for your answers. For scientific notation please use X.XXEX (i.e, 1230 as 1.23E3).
The first iteration of the exam is due 8-Mar-17. The answers will be posted on 9-Mar-17. Any incorrect answers can be resubmitted by 10-Mar-17. Changed answers will be worth 50 % of the original grade. For the 2nd resubmission the work related to the changes must be shown. Please post all questions to the blog.
There is an in-class meeting on the exam scheduled for Monday 6 March at 3:20 PM in Bagley 303A.
Hi professor, I'm having a lot of trouble with question 1 which is the only problem I have yet to answer. I seem to be drawing a blank and I can't figure out how to determine the relative concentrations based on pH. Do you have any suggestions to get started? Thank you.
ReplyDeleteThere is an example on: https://en.wikipedia.org/wiki/Stability_constants_of_complexes
Deletelook for the section on hydrolysis constants. This will provide some insight into the question. I will followup with more detail. I can also Canvas Conference on this if you wish.
You have K from the reaction. logK = -5.64
ReplyDeleteH2O + PuO2 2+ <----> PuO2OH+ + H+
For simplicity PuO2 2+ = M2+ and PuO2OH+ = MOH+
K = [MOH+][H+]/[M2+], ignore water
Kw = [OH-][H+] =1.01E-14
One needs to solve for the reaction
M2+ + OH- <----> MOH+
B= [MOH+]/[M2+][OH-]
So B and K have a common term, [MOH+]/[M2+]
Substitute K into the equation for B
K/[H+] = [MOH+]/[M2+]
B = [MOH+]/[M2+][OH-] =K/[H+][OH-]
with Kw = [H+][OH-]
B = K/Kw
One now has the value for B
At a selected pH, [OH-] can be found from pH + pOH =pKw
At any given pH one can calculate
[MOH+]/[M2+] = B [OH-]
From the question you have the sum of the relative concentrations is 1, so
[MOH+] + [M2+] = 1
One can use these two equations to solve for the terms [MOH+] and [M2+]
Thank you so much! viewed and submitted
ReplyDeleteExam handed in!
ReplyDeleteExam 3 has been submitted
ReplyDeleteExam 3 submitted
ReplyDeleteExam 3 has been submitted via email.
ReplyDeleteExam 3 submitted on Canvas.
ReplyDeleteExam 3 has been submitted via email.
ReplyDeletethanks for the exam 3 submissions
ReplyDeleteHi Ken,
ReplyDeleteI did question 3.1 wrong but I am not sure how. I used the Lambert-Beer law with extinction (I/I0) = 0.65, extinction coefficient 410 and length = 1cm. Could you maybe give a tip to the correct solution?
From the question A=0.35. So Beer's law
DeleteA=ebc, A=0.35, e=410.
Let me know if that helps, it should.
Thank you, I see what I did wrong now!
DeleteExam 3 resubmission has been sent via email.
ReplyDelete