Horvath Practice Problems 79

# Horvath Practice Problems 79 - K b 2 which is K b 2 = K w K...

This preview shows page 1. Sign up to view the full content.

Thus the concentration of HC 6 H 6 O 6 is 0 . 0025 mol HC 6 H 6 O 6 0 . 050 L = 0 . 05 M HC 6 H 6 O 6 This HC 6 H 6 O 6 then goes back towards the acid via the conju- gate base reaction, HC 6 H 6 O 6 + H 2 O ­ H 2 C 6 H 6 O 6 + OH So we need the K b 1 which is K b 1 = K w K a 1 = 1 . 27 × 10 10 The ICE chart is £ HC 6 H 6 O 6 ¤ £ OH ¤ [ H 2 C 6 H 6 O 6 ] I 0 . 05 0 0 C x x x E 0 . 05 x x x Using the K b 1 equation and solving for x = £ OH ¤ = 2 . 52 × 10 6 . Thus, pH = 14 ¡ log £ 2 . 52 × 10 6 ¤¢ = 8 . 40 Now, for the second equivalence point we repeat the procedure. The second equivalence point occurs at the addition of a total of 50 mL of NaOH. So, £ C 6 H 6 O 6 ¤ = 0 . 0025 mol 0 . 075 L = 0 . 033 M. Now we need the
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: K b 2 which is K b 2 = K w K a 2 = 6 . 25 × 10 − 3 The appropriate ICE chart for the reaction C 6 H 6 O 2 6 + H 2 O ­ HC 6 H 6 O − 6 + OH − is £ C 6 H 6 O − 6 ¤ £ OH − ¤ £ HC 6 H 6 O − 6 ¤ I . 033 C − x x x E . 033 − x x x Using the K b 2 equation and solving for x we get x = £ OH − ¤ = . 0116 . Thus pH = 14 − ( − log [0 . 0116]) 12 . 06 78 TITRATIONS PART II...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern