Lab 15 report

Gurkaran Singh Tomoko Hagane-Mullins May 1, 2022 Lab 15: Titration Curves “Drop the base” Part 1: A Mysterious Acid Titration Data Table: Volume of 0.200 M NaOH, mL First derivative pH 0 0.39 2.42 1 0.27 2.81 2 0.2 3.08 3 0.12 3.28 4 0.1 3.4 5 0.08 3.5 6 0.09 3.58 7 0.07 3.67 8 0.07 3.74 9 0.08 3.81 10 0.06 3.89 11 0.07 3.95 12 0.07 4.02 13 0.06 4.09 14 0.05 4.15 15 0.06 4.2 16 0.07 4.26 17 0.06 4.33 18 0.06 4.39 19 0.06 4.45 20 0.05 4.51 21 0.06 4.56 22 0.05 4.62

23 0.05 4.67 24 0.09 4.72 25 0.14 4.81 26 0.16 4.95 27 0.26 5.11 28 0.17 5.37 29 0.3 5.54 29.5 0.44 5.69 30 0.76 5.91 30.25 1.56 6.1 30.5 8.12 6.49 30.75 9.92 8.52 31 0.65 11 32 0.28 11.65 33 0.26 11.93 34 0.1 12.19 35 0.12 12.29 36 0.02 12.41 37 0.01 12.43 38 0.02 12.44 39 0.02 12.46 40 0.02 12.48 41 0.01 12.5 42 0.02 12.51 43 0.01 12.53 44 0.02 12.54 45 0.01 12.56 46 0.01 12.57 47 0.01 12.58 48 0.02 12.59 49 0.01 12.61

pH= 6.31 x 10^-5 Benzoic acid Name: Benzoic Acid. Red:1, Yellow: 0, Blue: 2 Chemical Formula: H ❑ C 7 H 5 O 2 Molar Mass: 122.13 g/mol Concentration of Weak Acid: Method 1: 3/122.13 g/0.10L =0.246 M Method 2: M=(Mb)(Vb)/(Va) =(0.200M)(30.75mL)/25L=0.246M Part 2: Polyprotic Acids Acid Dissociations: C 6 H 8 O 7 + H 2 O C ⇌ 6 H 7 O 7 - + H 3 O + Ka1 = 7.5 × 10 -4 C 6 H 7 O 7 - + H 2 O C ⇌ 6 H 6 O 7 2- + H 3 O + Ka2 = 1.7 × 10 -5 C 6 H 6 O 7 -2 + H 2 O C ⇌ 6 H 5 O 7 3- + H 3 O + Ka3 = 4.0 x 10 -7 Initial pH: Ka1 = 7.5 × 10 -4 7.5 × 10 -4 = x 2 0.50 − x x = 0.0189935

pH= -log ( 0.0189935) = 1.72 pH: 1.72 Midpoints: MP1: pH= -log ( 7.5 x 10 -4 ) = 3.125 MP2: pH= -log( 1.7 x 10 -5 ) = 4.77= 4.78 MP3: pH= -log( 4.0 x 10 -7 ) = 6.398= 6.40 Equivalence Points: Total volume = 100mL acid + 300mL NaOH = 400mL pH = pK a 1 + pK a 2 2 1 st Equivalence point : 3.12 + 4.78 2 = 3.95 2nd Equivalence point: 4.78 + 6.40 2 = 5.59 3rd Equivalence point: K w K a 3 = 1 × 10 − 14 4.0 × 10 − 7 = 2.5 × 10 -8 [A 3- ] = 0.5 × 0.1 L 0.4 L = 0.125 Kb = 2.5 × 10 -8 = x 2 0.125 − x x= 5.58892 × 10 -5 pOH = − ¿ log ( 5.58 × 10 -5 ) = 4.25 pH = 14 − ¿ 4.25 = 9.75 Past the Equivalence point : citric acid moles = 0.1L × 0.5M = 0.05mol Total volume = 100mL acid + 400mL NaOH = 500mL NaOH moles = 0.4L × 0.5M= 0.2 mol 0.2mol of NaOH − ¿ 0.05mol of citric acid = 0.15 mol M: 0.15 mol 0.1 L + 0.4 L = 0.30M pOH = − ¿ log (0.30M) = 0.53 pH= 14 − ¿ 0.53 = 13.47 Data Table: Volume of NaOH added pH

At 2nd Equivalence point → ❑ O ¿ 2 − ¿ C 6 H 6 ¿ At 3rd Midpoint → ❑ O ¿ 2 − ¿ C 6 H 6 ¿ and ❑ O ¿ 3 − ¿ C 6 H 5 ¿ At 3rd Equivalence point → ❑ O ¿ 3 − ¿ C 6 H 5 ¿ After 3rd Equivalence point → ❑ O ¿ 3 − ¿ C 6 H 5 ¿ References: - Smeureanu, G. & Geggier, S. (2019). General Chemistry Laboratory. New York, NY Zumdahl, S. (2014). Chemistry. 9th ed. Belmont, California: Cengage Learning Focus Question: Part1: Which substance is the crystalline powder? Benzoic acid substance is the crystalline powder. Part 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?) Before titration: Initial pH → C 6 H 8 O 7 1st Midpoint → C 6 H 8 O 7 and ❑ O ¿ − ¿ C 6 H 7 ¿ 1st Equivalence point → ❑ O ¿ − ¿ C 6 H 7 ¿ 2nd Midpoint → ❑ O ¿ − ¿ C 6 H 7 ¿ and ❑ O ¿ 2 − ¿ C 6 H 6 ¿ 2nd Equivalence point → ❑ O ¿ 2 − ¿ C 6 H 6 ¿ 3rd Midpoint → ❑ O ¿ 2 − ¿ C 6 H 6 ¿ and ❑ O ¿ 3 − ¿ C 6 H 5 ¿ 3rd Equivalence point → ❑ O ¿ 3 − ¿ C 6 H 5 ¿

After 3rd Equivalence point → ❑ O ¿ 3 − ¿ C 6 H 5 ¿ and OH - Post-Lab Assessment Questions: Part 1: 1. What other methods could you have used to identify the unknown chemical? By using calorimetry to measure the quantity of heat exchanged. IN addition, performing titration and pH indicator. 2. What is/are the dominant species in solution at pH = 4.7? The dominant species in solution at pH= 4.7 are H2C6H5O7^-, HC6H5O7^2-,H^+, and OH^-. 3. Does the solution at pH = 4.9 constitute a buffer? What about at pH= 5.9? Explain. The solution at pH = 4.9 constitutes a buffer because it fall under the buffer range of 3.2 to 5.2 pH. Unlike ph =5.9 does not constitute a buffer because the pH=5.9 is higher than the buffer range of 3.2 to 5.9. 4. How many moles of strong base can be added to the solution at pH = 4.7 before the buffer is exhausted? pH= 4.7 , volume = 24mL 27mL - 24mL = 3mL 0.003L × 0.200M = 0.0006 moles 5. The fire diamond for sodium azide looks like this: What does this tell you about the hazards this compound poses? What is this compound in the auto industry used for? Sodium azide is best known as the chemical found in automobile airbags. Red is 1 which is a flammability hazard. It tells us that sodium azide is slightly combustible and requires strong heat to ignite. Yellow is 3 which is an instability hazard. It tells us that sodium azide may detonate if shocked or heated under confinement or mixed with water. Blue is 4 which is Health hazard. It tells us that the sodium azide is highly toxic and can be fatal on short term exposure. 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 H20) in the solution? The major species at a pH of 6.0 is sodium citrate. 4 1 3