problem_set_2_key

problem_set_2_key - General Chemistry III, Chem-13 Problem...

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General Chemistry III, Chem-13 Problem Set II Key 1. Lactic acid is found in sour milk, in sauerkraut, and in muscles after activity. The K a for lactic acid is 1.4 × 10 -4 . a. If 2.75 grams of sodium lactate (NaCH 3 CHOHCO 2 ), is added to 5.00 × 10 2 ml of 0.100 M lactic acid (HCH 3 CHOHCO 2 ), what is the pH of the resulting buffer solution? Molar mass of sodium lactate = 112 grams/mol Molarity of sodium lactate = L mol gram grams 5 . 0 / 112 75 . 2 = 0.0491 M pH = pK a + log 55 . 3 100 . 0 0491 . 0 log 85 . 3 ] [ ] [ = + = Acid Base b. Is the final pH lower or higher than the pH of lactic acid solution? HA H + + 0.100 0 0 -x x x 0.100-x x x 4 2 10 4 . 1 100 . 0 - × = - = x x K a Solve for x = 0.00374 M and so pH = 2.42 The pH of the buffer is higher than that of the lactic acid.
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2. You dissolve 0.425 grams of NaOH in 2.00 L of a buffer solution that has [H 2 PO 4 ¯] = [HPO 4 2 ¯] = 0.132 M. a. What is the pH of the solution before adding the NaOH? H 2 PO 4 ¯ HPO 4 2 ¯ + H + K a2 = 6.2 × 10 -8 pH = pK a + log 20 . 7 132 . 0 132 . 0 log 20 . 7 ] [ ] [ = + = Acid Base b. What is the pH of the solution after adding the NaOH? Moles of NaOH added = 0.425/40 = 0.0106 The OH¯ ions from NaOH added will react with the acid part of the buffer which is the H 2 PO 4 ¯ ions. H 2 PO 4 ¯ + OH¯ H 2 O + HPO 4 2 ¯ Initial moles of H 2 PO 4 ¯ = 0.132 M × 2.00 L = 0.264 moles Initial moles of HPO 4 2 ¯ = 0.132 M × 2.00 L = 0.264 moles H 2 PO 4 ¯ + OH¯ H 2 O + HPO 4 2 ¯ Initial moles: 0.264 0.0106 0.264 Amount reacted: -0.0106 -0.0106 +0.0106 Amount left: 0.2534 0 0.2746 Concentrations: 0.130 M 0.137 M pH = pK a + log 22 . 7 130 . 0 137 . 0 log 20 . 7 ] [ ] [ = + = Acid Base
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3. What will the pH change be when 20.0 ml of 0.100 M NaOH is added to 80.0 ml of a buffer solution consisting of 0.169 M ammonia and 0.183 M ammonium chloride? K b for ammonia = 1.8 × 10 -5 (from Appendix D) K a = 10 5 14 10 6 . 5 10 8 . 1 10 0 . 1 - - - × = × × pK a = 9.26 pH = pK a + log 23 . 9 183 . 0 169 . 0 log 26 . 9 ] [ ] [ = + = Acid Base moles of NaOH added = 0.100 M × 0.020 L = 0.0020 moles The added NaOH will react with the acid of the buffer which is ammonium ions from the ammonium chloride. OH¯ + NH 4 + H 2 O + NH 3 Initial moles of NH 4 + = 0.183 M × 0.080 L = 0.01464 Initial moles of NH 3 = 0.169 M × 0.080 L = 0.01352 OH¯ + NH 4 + H 2 O + NH 3 Initial moles: 0.0020 0.01464 0.01352 Amount reacted: -0.0020 -0.0020 +0.0020 Amount left: 0 0.01264 0.01552 Concentration: M 126 . 0 1 . 0 01264 . 0 = M 155 . 0 1 . 0 01552 . 0 = pH = pK a + log 34 . 9 126 . 0 155 . 0 log 26 . 9 ] [ ] [ = + = Acid Base Change in pH = 9.26 – 9.34 = -0.08 Moral of the story: Addition of 20.0 ml of 0.100 M NaOH changes the pH only by eight hundredths of a pH unit.
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4. Phenol is a weak organic acid with the formula C 6 H 5 OH and has a K a of 1.3 × 10 -10 . Suppose 0.515 grams of this compound is dissolved in exactly 125 ml of water and
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problem_set_2_key - General Chemistry III, Chem-13 Problem...

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