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Unformatted text preview: 2/8/2010 Announcements: Monday, February 8, 2010
Homework WebAssign #5- Due Wednesday, February 10 at 8:00 AM Recitation Week 5 – “Buffers and Titrations” Lab 3- Titrations Complete prelab prior to recitation. Suggested Textbook Questions Continue Chapter 19 problems. Exam I Information Thursday, February 11, 2010
•Time and Place: 7:30 pm, Coors Events Center •Arrive between 7:00-7:10 pm •Enter through the northwest entrance of the Events Center •Pick up a lapboard as you enter •Sit with your lab group in assigned area •Bring NON-PROGRAMMABLE calculator, #2 pencil, eraser •No cell phones or headphones Format: 20 multiple choice questions, 100 points, 80 minutes Help Session: Wednesday, February 10, 5 pm, CHEM 140 Format: Instructor will respond to student questions Posted on CULearn: Exam I Information Sheet Two practice exams and keys Study Sheet February 8, 2010 - CT#1
Which of the following indicators would be best for titrating HCl with NaOH? Strong Acid/Strong Base Titration A. B. C. D. E. Crystal violet Bromphenol blue Bromcresol green Phenolphthalein Phenolphthalein Alizarin yellow R Weak Acid/Strong Base Titration Ex. 20.0 mL of 0.250 M HF (pKa = 3.17) with 0.100 M NaOH. 1) What is the equivalence volume? (Veq) 20.0 mL x 0.250 mmol HF x 1 mmol NaOH x mL NaOH soln = 50.0 mL XXXXXXXXXXmL HF solnXXXX1 mmol HFXXX 0.100 M NaOH 3) What is the pH when 25.0 mL of titrant have been added? (Vtitrant = ½Veq = 25.0 mL = Midpoint!) 20.0 mL x 0.250 M HF = 5.00 mmol HF 25.0 mL x 0.100 M NaOH = 2.50 mmol OH- 2) What is pH before any titrant is added? (Vtitrant =0 mL) pH of a 0.250 M HF solution. Just a weak acid solution! Use the Ka: Ka = x2/(0.250 – x) = 10-3.17 = x2/(0.250 – x) After the quadratic eqn: x = [H3O+] = 0.0127 M, pH = 1.897 Write out the reaction table: HF + OH F+ H2O I: 0 mmol 5 mmol 2.5 mmol -2.5 mmol +2.5 mmol C: -2.5 mmol E: 2.5 mmol 0 mmol 2.5 mmol Plug these numbers into H-H eqn. pH = 3.17 + log pH = pKa at the midpoint of a titration!
2.50 mmol F2.50 mmol HF pH = 3.17 1 2/8/2010 February 8, 2010 - CT#2
Which of the following is true for the pH at the equivalence point of any weak acid/strong base titration? 4) What is the pH after 50.0 mL of titrant have been added? (Vtitrant = Veq = 50.0 mL = Equivalence Point!)
20.0 mL x 0.250 M HF = 5.00 mmol HF 50.0 mL x 0.100 M NaOH = 5.00 mmol OHWrite out the reaction table: HF + OH- F0 mmol +5 mmol 5 mmol + H 2O A. pH < 7 B. pH = 7 HA + OH- → A- + H2O All HA has been converted to AA- is a weak base so the pH will be basic! I: C: E: 5 mmol -5 mmol 0 mmol 5 mmol -5 mmol 0 mmol C. pH > 7 C. pH
D. There is not enough information given. Not a buffer anymore! Just a weak base solution. Use Kb to solve! Vtotal = 20.0 mL + 50.0 mL = 70.0 mL [F-] = 5.00 mmol F-/70.0 mL = 0.0714 M Kb = 10(3.17 – 14) = 10-10.83 = x2/(0.0714 – x) = x2/0.0714 x = [OH-] = 1.03 x 10-6 pOH = 5.99 & pH = 8.01 5) What is the pH after 51.0 mL of titrant have been added? (Past Equivalence Point!)
20.0 mL x 0.250 M HF = 5.00 mmol HF 51.0 mL x 0.100 M NaOH = 5.10 mmol OHWrite out the reaction table: HF + OH- I: C: E: 5 mmol -5 mmol 0 mmol 5.1 mmol -5 mmol 0.1 mmol F+ H 2O Buffer Region (Use H-H Eqn) Weak Acid/Strong Base Titration
Excess strong base (get [OH-]) 0 mmol +5 mmol 5 mmol Midpoint HA sol’n (Use Ka) Equivalence Point A- sol’n (Use Kb) Excess strong base! Just a strong base solution now. Vtotal = 20.0 mL + 51.0 mL = 71.0 mL [OH-] = 0.10 mmol OH-/71.0 mL = 0.00141 M pOH = 2.85 & pH = 11.15 A- sol’n (Use Kb) Weak Base/Strong AcidTitration Midpoint Equivalence Point HA sol’n (Use Ka) Buffer Region (Use H-H Eqn) Excess strong acid (get [H3O+]) 2 ...
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This note was uploaded on 03/23/2010 for the course CHEM 1131 taught by Professor Redin,kend during the Spring '08 term at Colorado.
- Spring '08