Homework 07 Solutions

# Homework 07 Solutions - Gadhia Tejas Homework 7 Due 11:00...

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Gadhia, Tejas – 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 cyanic acid (HOCN) 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 cyanic acid. C a = 0 . 66, C b = 0 . 528, and the volume of HOCN is 100 mL. 1. 8 . 23 2. None of these 3. 3 . 5 correct 4. 80 5. 40 Explanation: K a = 3 . 5 × 10 - 4 C a = 0 . 66 C b = 0 . 528 V HOCN = 100 mL 0 2 4 6 8 10 12 14 0 20 40 60 80 100 120 140 Volume of base (mL) pH (80,8 . 23) (40,3 . 5) The equivalence point of this titration is when the curve is at an inflection point (nearly vertical); i.e. , at a volume of 80 mL . The pH at the equivalence point of this titration is 8 . 23 pH . The p K a can be found at one-half the vol- ume of the equivalence point; i.e. , at 40 mL. The p K a is 3 . 5 pH from looking at the graph. The formula is p K a = - log ( K a ) = - log 3 . 5 × 10 - 4 · = 3 . 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. 0.200 M 2. 1.11 M 3. 0.555 M 4. 0.278 M correct Explanation: V NaOH = 25.0 mL [NaOH] = 0.333 M

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