{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

3304%20Homework%20#2%20%28chapter%202%29

3304%20Homework%20#2%20%28chapter%202%29 - Homework#2 BCHS...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Homework #2 BCHS 3304 Read FOB chapter #2 Physical properties of water p22- 36. Read and Review Students companion Chapter 2 water p 11-14. Questions 1. How many milligrams of cytochrome c (MW=12,500) would you need to make a 10 ml of a 50 μ M solution? 2. What is the molar concentration of pure glycerol? The density of glycerol is 1.26 g/cm 3 . 2. 13.68 M 3. Write a short one-line definition for the Key Terms on page 37. You are responsible for these terms, but do not turn them in. 4. Problems in FOB, pg 37. 1, 8, 10, 13, 14 and 18. 8. Indicate the ionic species that predominates at pH 4, 8, and 11 for
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
a) Ammonia: pH4 NH4+ / pH8 NH4+ / pH 11 NH3 b) Phosphoric acid: pH4 H2PO4- / pH8 HPO4 2- / pH 11 HPO4 2- 10. Calculate the pH of a 1 L solution containing *pH= -log [H+] [H+]= Kw/[OH-] Kw (a constant)= 10^-14 M^2 a) 10 mL of 5 M NaOH (0.010 L)(5 mol*L^-1 NaOH)/(1L)= 0.05 M NaOh or 0.05 OH- [H+]= (10^-14)/(0.05) =2*10^-13 M pH= -log(2*10^-13 M) = 12.7 b) 10 mL of 100 mM glycine and 20 mL of 5 M HCl (0.020 L)(5 mol*L^-1 HCl)/(1L)= 0.1 M HCl = 0.1 M H+ pH= -log(0.1)= 1.0 c) 10 mL of 2 M acetic acid and a 5 g of sodium acetate (MW: 82 g*mol^-1) pH= pK+ log([acetate]/[acetic acid]) [acetate]= (5g)(1 mol/82 g)/(1 L) = 0.061 M
Image of page 2
[acetic acid]= (0.01 L)(2 mol)/(1L) = 0.02 M
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern