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problem_set_3_key - General Chemistry III Chem-13 Problem...

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General Chemistry III, Chem-13 Problem Set III Key 1. What volume of 0.275 M potassium hydroxide must be added to 75.0 ml of 0.137 M H 3 AsO 4 solution to reach: The K a values for H 3 AsO 4 can be found in Appendix D K a1 = 5.6 × 10 -3 pK a1 = 2.25 K a2 = 1.0 × 10 -7 pK a2 = 7.00 K a3 = 3.0 × 10 -12 pK a2 = 11.52 Milli Moles of H 3 AsO 4 = 75.0 ml × 0.137 M = 10.275 a. the first equivalence point moles of H 3 AsO 4 = moles of OH¯ H 3 AsO 4 + OH¯ H 2 AsO 4 ¯ + H 2 O Starting moles 10.275 10.275 Molarity of OH¯ = 0.275 M Volume of OH¯ = milli-moles/Molarity = 10.275/0.275 = 37.4 ml b. the second equivalence point milli-moles of OH¯ at the second equivalence point = 2 × milli-moles of H 3 AsO 4 Therefore volume at second equivalence point = 2 × 37.36 = 74.7 ml c. the third equivalence point milli-moles of OH¯ at the third equivalence point = 3 × milli-moles of H 3 AsO 4 Therefore volume at second equivalence point = 3 × 37.36 = 112 ml
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2. You are titrating 10.0 ml of a 0.1 M solution of the fully protonated form of the amino acid glycine (H 2 G + ) whose K a1 is 4.47 × 10 -3 and K a2 is 1.66 × 10 -10 with 0.1 M sodium hydroxide. Draw a detailed titration curve and label all the parts of the curve. From the table of indicators that was given to you in class, pick the appropriate indicators that you would use for this titration. First lets calculate the starting pH: H 2 G + H + + HG 0.1 – x x x 3 2 1 10 47 . 4 1 . 0 - × = - = x x K a x = 0.021 pH = 1.67 Next lets calculate the volume of NaOH for the equivalence points: Milli-moles of glycine = 10.0 ml × 0.1 M = 1.0 Therefore milli-moles of NaOH at the first equivalence point = 1.0 Since the molarity of NaOH = 0.1 M, the volume of NaOH for the first equivalence
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