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CH 302 HW7 - Thompson Margaret Homework 7 Due 11:00 pm Inst...

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Thompson, Margaret – 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 hypochlorous acid (HOCl) with NaOH(aq) base is given below. 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 hypochlorous acid. C a = 0 . 56, C b = 0 . 5544, and the volume of HOCl is 100 mL. 1. 49 . 5 2. 99 3. 7 . 5 correct 4. 10 . 2 5. None of these Explanation: K a = 3 . 5 × 10 - 8 C a = 0 . 56 C b = 0 . 5544 V HOCl = 100 mL 0 2 4 6 8 10 12 14 0 20 40 60 80 100 120 140 Volume of base (mL) pH (99,10 . 2) (49 . 5,7 . 5) The equivalence point of this titration is when the curve is at an inflection point (nearly vertical); i.e. , at a volume of 99 mL . The pH at the equivalence point of this titration is 10 . 2 pH . The p K a can be found at one-half the vol- ume of the equivalence point; i.e. , at 49 . 5 mL. The p K a is 7 . 5 pH from looking at the graph. The formula is p K a = - log ( K a ) = - log 3 . 5 × 10 - 8 · = 7 . 45593 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. 0.555 M 3. 0.278 M correct 4. 0.200 M Explanation: V NaOH = 25.0 mL [NaOH] = 0.333 M
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Thompson, Margaret – 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: 8 . 2 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) (18 . 5, 4 . 8) (37, 8 . 2) 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: 37 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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