I ii volume of base added ph the ph at point i is

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I II Volume of base added pH The pH at point I is equal to the and the pH at point II is 7. 1. pH of the acid; less than 2. p K a of the acid; greater than correct 3. p K a of the acid; equal to 4. pH of the acid; greater than 5. p K a of the acid; less than Explanation: 025(part1of2)3.4points The titration curve for the titration of 0 . 73 M H 2 SO 3 (aq) with 0 . 73 M KOH(aq) is given below. 0 2 4 6 8 10 12 14 0 20 40 60 80 100 120 140 Volume of base (mL) pH
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casey (rmc2555) – Homework 9 – holcombe – (51395) 9 What are the major species in solution after 100 mL of KOH(aq) have been added? 1. HSO - 3 (aq), SO 2 - 3 (aq), and K + (aq) 2. SO 2 - 3 (aq), OH - (aq), and K + (aq) 3. H 2 SO 3 (aq), HSO - 3 , and K + (aq) 4. HSO - 3 (aq) and K + (aq) 5. SO 2 - 3 (aq) and K + (aq) correct Explanation: 0 2 4 6 8 10 12 14 0 20 40 60 80 100 120 140 Volume of base (mL) pH (100,10 . 3) (25,1 . 9) (75,7 . 2) 026(part2of2)3.4points What are the major species in solution after 75 mL of KOH(aq) have been added? 1. SO 2 - 3 (aq), and K + (aq) 2. SO 2 - 3 (aq), OH - (aq), and K + (aq) 3. HSO - 3 (aq), SO 2 - 3 (aq), and K + (aq) cor- rect 4. H 2 SO 3 (aq), HSO - 3 , and K + (aq) 5. HSO - 3 (aq) and K + (aq) Explanation: 027 3.4points You have a solution that is buffered at pH = 2.0 using H 3 PO 4 and H 2 PO - 4 (p K a1 = 2 . 12; p K a2 = 7 . 21; p K a3 = 12 . 68). You decide to titrate this buffer with a strong base. 15.0 mL are needed to reach the first equivalence point. What is the total volume of base that will have been added when the second equivalence point is reached? 1. 30 mL 2. A second equivalence point in the titra- tion will never be observed. 3. > 30 mL correct 4. < 30 mL Explanation: 028 3.4points For the titration of 50.0 mL of 0.020 M aque- ous salicylic acid with 0.020 M KOH(aq), cal- culate the pH after the addition of 55.0 mL of KOH(aq). For salycylic acid, p K a = 2.97. 1. 10.98 correct 2. 12.30 3. 7.00 4. 12.02 5. 11.26 Explanation: 029 3.4points What will be the pH of the solution after 200 mL of 0.3 M NaOH is added to 100 mL of 0.6 M HClO ( K a = 2 × 10 - 7 )? 1. 8 2. 7 3. 11 4. 10 correct
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casey (rmc2555) – Homework 9 – holcombe – (51395) 10 Explanation: The final volume is three times the initial volume, so the final concentration is 0.2 M. All of the HClO has been converted to its conjugate, so the pH is given by a simple weak acid/base calculation. 030 3.4points What is the volume of 0.208 M NaOH required to titrate 12.6 mL of 0.280 M HF to the equivalence point? 1. 0.170 L 2. 1.70 L 3. 0.0170 L correct 4. 1.70 cm 3 Explanation: V HF = 12.6 mL [HF] = 0.28 M [NaOH] = 0.208 M NaOH + HF Na + + F - + H 2 O 12 . 6 mL HF(aq) × 0 . 280 mol HF 1000 mL HF(aq) × 1 mol NaOH 1 mol HF × 1000 mL NaOH(aq) 0 . 208 mol NaOH = 16 . 9615 mL NaOH(aq) = 0 . 0169615 L
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