Lab Report Titration Curves “Drop the base”

Titration Curves “Drop the base” 1.0 Introduction In the first part of this lab, we identified a weak acid using data of a titration that was performed with the acid and 0.2 M NaOH. We graphed this data and using the graph we identified the buffer region, midpoint, and equivalence point of the titration. To get a more accurate reading of the volume of NaOH at the equivalence point we calculated and graphed the derivative, dpH over dV. With this more accurate reading, we were also able to get a more accurate value of the volume of NaOH at the midpoint. Then we got the pH at the midpoint, which is equal to the pKa. This allowed us to identify the weak acid as being Benzoic acid. In the second part of this lab, we calculated the pH of a titration of 100 mL od 0.50 M citric acid an d0.50 M NaOH, before NaOH was added, at the three midpoints, and three equivalence points, and past the equivalence point. Using these eight data point, we graphed a titration curve, and labeled all mid- and equivalence points, and identified the major species in the solution, at that point. 2.0 Materials - Lab Manual - Calculator 3.0 Observation and Experimental Part One: Volume 0.200 M NaOH, mL 0 1 2 3 4 5 6 7 8 9 10 11 12 13 pH 2.42 2.81 3.08 3.28 3.4 3.5 3.58 3.67 3.74 3.81 3.89 3.95 4.02 4.09 NaOH(dpH/dV) 0.39 0.27 0.2 0.12 0.1 0.08 0.09 0.07 0.07 0.08 0.06 0.07 0.07 0.06 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 4.15 4.2 4.26 4.33 4.39 4.45 4.51 4.56 4.62 4.67 4.72 4.81 4.95 5.11 5.37 0.05 0.06 0.07 0.06 0.06 0.06 0.05 0.06 0.05 0.05 0.09 0.14 0.16 0.26 0.17 29 29.5 30 30.25 30.5 30.75 31 32 33 34 35 36 37 5.54 5.69 5.91 6.1 6.49 8.52 11 11.65 11.93 12.19 12.29 12.41 12.43 0.3 0.44 0.76 1.56 8.12 9.92 0.65 0.28 0.26 0.1 0.12 0.02 0.01 38 39 40 41 42 43 44 45 46 47 48 49 50 12.44 12.46 12.48 12.5 12.51 12.53 12.54 12.56 12.57 12.58 12.59 12.61 12.62 pH vs. volume of NaOH 14 12 Buffer Zone pH 10 Equivalence Point 8 Midpoint 6 4 2 0 0 10 20 30 40 Volume of NaOH (mL) Equivalence Point: at 30.75 mL and pH = 8.52 Midpoint: ~15 mL and pH 4.2 pKa = 4.2 Buffer zone: 3.2 – 5.2 50 60 0.02 0.02 0.02 0.01 0.02 0.01 0.02 0.01 0.01 0.01 0.02 0.01 0.2524 NaOH(dpH/dV) vs. Volume NaOH 12 NaOH (dpH/dV) 10 8 6 4 2 0 0 10 20 30 40 50 60 Volume of NaOH (mL) Equivalence Point is at 30.75 mL of NaOH and a pH of 9.92 Midpoint is at 15.375 mL of NaOH and a pH of 4.20 pH = pKa = 4.20 Ka = 10−4.20 = 6.31 ∗ 10−5 Benzoic acid (HC7H5O2) Method 1: HC7 H5 O2 + NaOH ⇌ NaC7 H5 O2 + H2 O Volume of 0.200 M NaOH at Equivalence point was 30.75 mL 30.75 mL ∗ 1L ∗ 0.200M = 0.00615 mol of NaOH 1000 mL 1:1 ration of NaOH and acid 0.00615 mol of acid 1000 mL = 0.246 M Acid ∗ 1L 25 mL Method 2: Initial pH = 2.42 [H+] = 10-2.42 = 0.0038 M Ka = 6.31 ∗ 10−5 = X = 0.229 M Acid (0.0038)2 𝑥 Name: Benzoic Acid Chemical formula: HC7 H5 O2 1 Molar Mass: 122.118 g/mol 1 0 Part Two: H3 C6 H5 O7 + H2 O ⇌ H2 C6 H5 O−7 + H3 O+ Ka1 = 7.5*10-4 H2 C6 H5 O−7 + H2 O ⇌ HC6 H5 O7−2 + H3 O+ Ka2 = 1.7*10-5 −3 + HC6 H5 O−2 7 + H 2 O ⇌ C6 H 5 O 7 + H 3 O Ka3 = 4.0*10-7 Initial pH: H3 C6 H5 O7 H2 C6 H5 O−7 H 3 O+ 0.5 0 0 C -x +x +x E 0.5 - x x x I 7.5 ∗ 10−4 = 𝑥2 0.5 − 𝑥 𝑥 2 + 7.5 ∗ 10−4𝑥 − 3.75 ∗ 10−4 = 0 𝑥= −(7.5 ∗ 10−4) ± √(7.5 ∗ 10−4 )2 + 4(3.75 ∗ 10−4 ) 2 𝑥 = 1.899 ∗ 10−2 pH = − log(1.899 ∗ 10−2) = 1.72 Midpoint 1: pH = pKa1 = − log(7.5 ∗ 10−4 ) = 3.12 Midpoint 2: pH = pKa2 = − log(1.7 ∗ 10−5 ) = 4.77 Midpoint 3: pH = pKa3 = − log(4.0 ∗ 10−7 ) = 6.40 Equivalence Point 1: pH = 1 (3.12 + 4.77) = 3.95 2 pH = 1 (6.40 + 4.77) = 5.59 2 Equivalence Point 2: Equivalence Point 3: H3 C6 H5 O7 + OH− ⇌ H2 C6 H5 O7− + H2 O nH3 C6 H5 O7 = 0.5 M ∗ .1L = 0.05 mol At equivalence point nH3 C6 H5 O7 = nOH − = 0.05 mol Volume of NaOH = 0.05mol = 0.1 L ∗ 1000mL = 100mL 0.5M Volume added at Equivalence point 3 = 3 * 100 mL = 300 mL −2 − C6 H5 O−3 7 + H2 O ⇌ HC6 H5 O7 + OH 0.5𝑀 𝐶𝑖𝑡𝑟𝑖𝑐 𝑎𝑐𝑖𝑑 ∗ 0.1 𝐿 = 0.05 𝑚𝑜𝑙 0.05 𝑚𝑜𝑙 = 0.125 𝑀 C6 H5 O7−3 0.1 𝐿 + 0.3 𝐿 𝐾𝑏 = 10−14 = 2.5 ∗ 10−8 4.0 ∗ 10−7 C6 H5 O−3 7 H2 C6 H5 O−2 7 OH− 0.125 0 0 C -x +x +x E 0.125 - x x x I 2.5 ∗ 10−8 = 𝑥2 0.125−𝑥 (5% rule) x = 5.59*10-5 𝑝𝑂𝐻 = − log(5.59 ∗ 10−5) = 4.25 𝑝𝐻 = 14 − 4.25 = 9.75 Past Equivalence Point: n citric acid = 0.1 L ∗ 0.5 M = 0.05 mol nNaOH = 0.4 L ∗ 0.5 M = 0.2 mol nNaOH left over = 0.2 mol − 0.05 mol = 0.15 mol [NaOH] = 0.15 mol = 0.3 0.1 L + 0.4 L pOH = − log(0.3) = 0.523 pH = 14 − 0.523 = 13.48 Major Species pH vs. Volume of 0.50 M NaOH Initial: H3 C6 H5 O7 16 Midpoint 1: H3 C6 H5 O7 , H2 C6 H5 O−7 14 12 Midpoint 3 pH 10 Midpoint 2: H2 C6 H5 O−7, HC6 H5 O7−2 Midpoint 2 8 Equivalence Point 3 Midpoint 1 6 4 Midpoint 3: −3 HC6 H5 O−2 7 , C6 H5 O7 Equivalence Point 2 Equivalence Point 1 2 0 0 50 100 150 200 250 Equivalence Point 1: H2 C6 H5 O−7 300 350 Volume of NaOH (mL) 400 450 Equivalence Point 2: HC6 H5 O−2 7 Equivalence Point 3: C6 H5 O−3 7 Last: Excess OH- 4.0 Discussion and Conclusion In part one of this lab, we were able to identify the mysterious weak acid as benzoic acid, with the formula HC7 H5 O2 , and a molar mass of 122.118 g/mol. We also found that the health hazard and the flammability hazard was of this acid was slight, the instability hazard was minimal, and it did not have any specific hazards. Errors in both sections of this lab could have occurred from calculation, mainly from rounding. 5.0 Focus Questions 1. Which substance is the crystalline powder? (Part 1) The crystalline powder is Benzoic acid. 2. What are the major species in solution during the titration of citric acid with a strong base? (i.e. which form of citric acid predominates at each of the 8 data points) (Part 2) Water is always a major species. - Initial: H3 C6 H5 O7 Midpoint 1: H3 C6 H5 O7 , H2 C6 H5 O−7 Midpoint 2: H2 C6 H5 O−7, HC6 H5 O−2 7 −3 Midpoint 3: HC6 H5 O−2 7 , C6 H5 O7 − Equivalence Point 1: H2 C6 H5 O7 and NaOH Equivalence Point 2: HC6 H5 O−2 7 and NaOH Equivalence Point 3: C6 H5 O−3 7 and NaOH Past last equivalence point: Excess NaOH 6.0 References Smeureanu, Gabriela. “Experiment 15 Titration Curves ‘Drop the base’” Chemistry 106 General Chemistry Laboratory, edited by Stephanie Geggier, Custom Publishing, Inc., 2019, pp. 113123. 7.0 Post Lab Questions Part 1: 1. What other methods could you have used to identify the unknown chemical? We could have performed tests to determine physical properties of the substance. For example, solubility tests, or a test to se how flammable it is, or how and if it reacts under different conditions. 2. What is/are the dominant species in solution at pH = 4.0? At pH 4.0 the titration was at the mid-point which means that the major species were water, HC7H5O3, and C7H5O3-, 3. Does the solution at pH = 4.0 constitute a buffer? Explain. Yes, the pKa of benzoic acid is 4.2, which means that the buffer zone would be from pH 3.2 – 5.2, and 4.2 is between that. 4. How many moles of strong base can be added to the solution at pH = 4.0 before the buffer is exhausted? The data the was given to us shows that about 12 mL of NaOH was added to the solution to reach a pH of 4, and about 27.5 mL of NaOH was added to reach a pH of 5.2. mL of NaOH added to exhaust buffer = 27.5 mL − 12.0 mL = 15.5 mL 1L ∗ 0.200 𝑀 = 0.0031 mol 1000 mL 0.0031 of NaOH can be added at pH 4.0 before the buffer is exhausted. 15.5 mL ∗ 5. The fire diamond for sodium azide looks like this: a. What does this tell you about the hazards this 1 compound poses? This compound is highly toxic and could be fatal with 4 3 only minimal exposure. It is slightly combustible, which means it needs a lot of heating to ignite. It is also seriously unstable, which means it could detonate if heated under confinement, mixed with water, or shocked. b. What is this compound in the auto industry used for? Sodium azide is what makes airbag blow up during a car crash. Just a small electrical charge caused when a car crashed causes it to explode and converts into nitrogen bag. Part 2: 6. Equal volumes of 1 M solutions of citric acid, sodium citrate, sodium hydrogen citrate, and sodium dihydrogen citrate are combined, and the pH is adjusted to 6.0 using 1 M NaOH. What is/are the major species (other than water) in the solution? At pH 6.0 this solution is between the 2nd equivalence point and the 3rd midpoint, therefore −3 the major species would be HC6 H5 O−2 7 , and C6 H5 O7 .