HW 7 Solutions - Wright Jacqueline Homework 7 Due 11:00 pm...

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Wright, Jacqueline – Homework 7 – Due: Oct 29 2007, 11:00 pm – Inst: James Holcombe 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The curve for the titration of formic acid (HCOOH) with NaOH(aq) base is given be- low. 0 2 4 6 8 10 12 14 0 20 40 60 80 100 120 140 Volume of base (mL) pH Estimate the p K a of formic acid. C a = 0 . 66, C b = 0 . 6072, and the volume of HCOOH is 100 mL. 1. None of these 2. 46 3. 92 4. 3 . 7 correct 5. 8 . 37 Explanation: K a = 1 . 8 × 10 - 4 C a = 0 . 66 C b = 0 . 6072 V HCOOH = 100 mL 0 2 4 6 8 10 12 14 0 20 40 60 80 100 120 140 Volume of base (mL) pH (92,8 . 37) (46,3 . 7) The equivalence point of this titration is when the curve is at an inflection point (nearly vertical); i.e. , at a volume of 92 mL . The pH at the equivalence point of this titration is 8 . 37 pH . The p K a can be found at one-half the vol- ume of the equivalence point; i.e. , at 46 mL. The p K a is 3 . 7 pH from looking at the graph. The formula is p K a = - log ( K a ) = - log 1 . 8 × 10 - 4 · = 3 . 74473 pH . Note : The p K a is the pH when the mole fraction is 0.5. 002 (part 1 of 1) 10 points It required 25.0 mL of 0.333 M NaOH solution to completely neutralize 15.0 mL of H 2 SO 4 so- lution. What was the molarity of the H 2 SO 4 ? 1. 1.11 M 2. 0.200 M 3. 0.278 M correct 4. 0.555 M Explanation: V NaOH = 25.0 mL [NaOH] = 0.333 M
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Wright, Jacqueline – Homework 7 – Due: Oct 29 2007, 11:00 pm – Inst: James Holcombe 2 V H 2 SO 4 = 15.0 mL The balanced equation for the reaction is H 2 SO 4 + 2 NaOH Na 2 SO 4 + 2 H 2 O We determine the moles of NaOH used. ? mol NaOH = 0 . 025 L soln × 0 . 333 mol NaOH 1 L soln = 0 . 00832 mol NaOH Using the mole ratio from the balanced chemical equation we calculate the moles of H 2 SO 4 needed to react with this amount of NaOH: ? mol H 2 SO 4 = 0 . 00832 mol NaOH × 1 mol H 2 SO 4 2 mol NaOH = 0 . 00416 mol H 2 SO 4 This is the amount of H 2 SO 4 that must have been in the 15.0 mL sample. Molarity is moles solute per liter of solution: ? M H 2 SO 4 = 0 . 00416 mol H 2 SO 4 0 . 0150 L solution = 0 . 277 M H 2 SO 4 003 (part 1 of 3) 10 points Below is the pH curve of a weak acid (HA) titrated with strong base. Answer the fol- lowing question based on the interpretation of this pH curve. Be as accurate as possible. 0 5 10 15 20 25 30 35 40 45 50 55 60 0 1 2 3 4 5 6 7 8 9 10 11 12 13 Titration Curve Amount of Base added (mL) pH What is the pH at the equivalence point of this titration? Your answer must be within ± 10%. Correct answer: 7 . 6 pH. Explanation: The inflection points are shown below. 0 5 10 15 20 25 30 35 40 45 50 55 60 0 1 2 3 4 5 6 7 8 9 10 11 12 13 Titration Curve Amount of Base added (mL) Acid/Base concentration (pH) (14 . 5, 4) (29, 7 . 6) 004 (part 2 of 3) 10 points How much base much be added to make the solution equalized? Your answer must be within ± 10%. Correct answer: 29 mL. Explanation: 005 (part 3 of 3) 10 points What is the p K a for this acid? Your answer must be within ± 5%.
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