Titration Curves Drop the Base Lab Report

Lab Report # 15: Titration Curves ÒDrop the baseÓ Part I: A Mysterious Acid Volume 0.200 M NaOH, mL pH Derivative Volume 0.200 M NaOH, mL pH Derivative 0.00 2.42 N/A 19.00 4.45 0.06 1.00 2.81 0.40 20.00 4.51 0.06 2.00 3.08 0.27 21.00 4.56 0.05 3.00 3.28 0.20 22.00 4.62 0.06 4.00 3.40 0.12 23.00 4.67 0.05 5.00 3.50 0.10 24.00 4.72 0.05 6.00 3.58 0.08 25.00 4.81 0.09 7.00 8.00 3.67 3.74 0.09 0.07 26.00 27.00 4.95 5.11 0.14 0.16 9.00 3.81 0.07 28.00 5.37 0.26 10.00 3.89 0.08 29.00 5.54 0.17 11.00 3.95 0.06 29.50 5.69 0.30 12.00 4.02 0.07 30.00 5.91 0.44 13.00 4.09 0.07 30.25 6.10 0.76 14.00 4.15 0.06 30.50 6.49 1.56 15.00 4.20 0.05 30.75 8.52 8.12 16.00 4.26 0.06 31.00 11.00 9.92 17.00 4.33 0.07 32.00 11.65 0.65 18.00 4.39 0.06 33.00 11.93 0.28 Volume 0.200 M NaOH, mL pH Derivative 34.00 12.19 0.26 35.00 12.29 0.10 36.00 12.41 0.12 37.00 12.43 0.02 38.00 12.44 0.01 39.00 12.46 0.02 40.00 12.48 0.02 41.00 12.50 0.02 42.00 12.51 0.01 43.00 12.53 0.02 44.00 12.54 0.01 45.00 12.56 0.02 46.00 12.57 0.01 47.00 12.58 0.01 48.00 12.59 0.01 49.00 12.61 0.02 50.00 12.62 0.01 At mid point [HA] = [A-] Thus pH = pKa = 4.20 Ka = 10-4.20 = 6.3 * 10-5 Benzoic acid HC7H7O2 (actual Ka 6.4 * 10-5) Concentration of acid Method 1: Equivalence point volume 30.75 mL NaOH (0.03075 L) * (0.200 M) = 0.00615 mol (0.00615 mol) / (0.025 L) = 0.246 M acid Method 2: Starting pH 2.42 [H+] = 10-2.42 = .0038 M Ka = 6.3 * 10-5 = (.0038)2/x x = 0.229 M Name: Benzoic Acid Chemical Formula: HC7H5O2 Molar Mass: 122.12 g/mol Focus Question: 1. The crystalline powder is Benzoic Acid. Part II: Polyprotic Acids Initial pH: Ka1 = [H3O+][C6H7O7-]/[C6H8O7] = 7.5 * 10-4 = x2/(0.5 - x) ≈ x2/0.5 x = 0.0194 pH = -log[H3O+] = -log(0.0194) = 1.71 Midpoint 1: pH = pKa1 = -log(Ka1) = -log(7.5 * 10-4) = 3.12 Midpoint 2: pH = pKa2 = -log(Ka2) = -log(1.7 * 10-5) = 4.77 Midpoint 3: pH = pKa3 = -log(Ka3) = -log(4 * 10-7) = 6.40 Equivalence 1: (pKa1 + pKa2)/2 = (3.12 + 4.77)/2 = 3.95 Equivalence 2: (pKa2 + pKa3)/2 = (4.77 + 6.40)/2 = 5.59 Equivalence 3: C6H8O7(aq) + NaOH(aq) → C6H5O72-(aq) + 3H2O(l) Volume of NaOH = (100 mL * 0.5 M * 3)/0.5 M = 300 mL 0.5 M citric acid * 0.100 L = 0.05 mol 0.05 mol/(0.1 L + 0.3 L) = 0.125 M 2.5 * 10-8 = x2/(0.125 - x) ≈ x2/0.125 x = 5.59 * 10-5 M pOH = -log[OH-] = -log(5.59 * 10-5) = 4.25 pH = 14 - pOH = 14 - 4.25 = 9.25 Volume 0.5 M NaOH, mL pH Past equivalence (400 mL of NaOH): 0.00 1.71 mol citric acid = (0.1 L) * (0.5 M) = 0.05 mol 50.00 3.12 mol NaOH = (0.4 L) * (0.5 M) = 0.2 mol 100.00 3.95 Difference = 0.2 - 0.05 = 0.15 mol 150.00 4.77 200.00 5.59 250.00 6.40 300.00 9.25 400.00 13.47 Concentration = (0.15 mol)/(0.1 L + 0.4 L) = 0.3 [NaOH] = [OH-] pOH = -log[OH-] = -log(0.3) = 0.53 pH = 14 - pOH = 14 - 0.53 = 13.47 Focus Question: 1. Water is always a major species and hydronium is a major species until the pH of 7. From the initial starting point till the Þrst midpoint, the major species was C6H8O7(aq). From the Þrst midpoint to the second midpoint, "C6H7O7-(aq) was the major species. From the second midpoint to the third midpoint, C6H6O72-(aq) was the major species. From the last midpoint on, "C6H5O73-(aq) was left was the major species. Part III: Post Lab Questions 1. Other methods we could have used to identify the unknown chemical is by performing a solubility test, chemical tests, ßammability tests, and measurements of physical properties. The solubility test would predict the polarity, size of the unknown chemical and the basic or acid functional groups. Chemical tests and ßammability tests can help distinguish inorganic and organic compounds of the unknown chemical. The measurements of physical properties will give you the density, boiling point, and melting point of the solution. 2. The dominant species in the solution at pH=4.5 is C6H7O7- ,C6H6O72-, [OH- ], and [H+ ]. 3. Yes, a pH of 4.5 does constitute a buffer because we approximated that the buffer range is from 3.08 to 5.08 and 4.5 falls in between that. In that region, the rate of change is 0, and that only happens in the buffer zone. 4. (0.200M)(0.0065L)= 0.0013 moles of NaOH can be added 5. This compound poses a level 4 health hazard: EXTREME --- Highly toxic Ð may be fatal on short term exposure. It also poses a level 1 ßammability hazard: SLIGHT --- Slightly combustable. Requires strong heating to ignite. Lastly, it poses a level 3 instability hazard: SERIOUS --- May detonate if shocked or heated under conÞnement or mixed water. White: none. Sodium azide is converted to nitrogen oxide inside the airbag. It is the component in airbags that explodes when an electric charge is triggered through a car impact. 6. The major species in the solution are sodium citrate, sodium hydrogen-citrate, [OH-], and [H+] are present at a pH of 9.5.