Free Response 2001 C Acids and Bases

1. Free Response 2001 CAcids and Bases Barb Fallon AP Chemistry June 2007

2. The Questions: A, B, and C • A. The amount of acetylsalicylic acid (ASA) in a single aspirin tablet is 325 mg, yet the tablet has a mass of 2.00 g. Calculate the mass percent of acetylsalicylic acid in the tablet. • B. The elements contained in ASA are hydrogen, carbon, and oxygen. The combustion of 3.000 g of the pure compound yields 1.200 g of water and 3.72 L of dry carbon dioxide, measure at 750. mmHg and 25 C. Calculate the mass, in g, of each element in the 3.000 g sample. • C. A student dissolved 1.625 g of pure ASA in distilled water and titrated the resulting solution to the equivalence point using 88.43 mL of 0.102 M NaOH(aq). Assuming that ASA has only one ionizable hydrogen, calculate the molar mass of the acid.

3. The Questions: D • D. A 2.00 x 10-3 mole sample of pure ASA was dissolved in 15.00 mL of water and then titrated with 0.100 M NaOH(aq). The equivalence point was reached after 20.00 mL of the NaOH solution had been added. Using the data from the titration, shown in the table below, determine: • (i) the value of the acid dissociation constant, pKa, for ASA • (ii) the pH of the solution after a total volume of 25.00 mL of the NaOH solution had been added (assume that volumes are additive).

4. Part A: Mass Percent • A. The amount of acetylsalicylic acid (ASA) in a single aspirin tablet is 325 mg, yet the tablet has a mass of 2.00 g. Calculate the mass percent of acetylsalicylic acid in the tablet. • So, this question is straightforward: find the mass percent of ASA in the tablet.

5. Part A: Mass Percent • Mass percent of compound X is defined as (mass of compound X) / (total mass) x 100%. • Mass percent of ASA = (mass of ASA in sample) / (total mass of sample) x 100%. • Mass percent ASA = (.325 g ASA) / (2.00 g) x 100% = 16.25% • Answer: the mass percent of ASA in the tablet is 16.25%.

6. Part B: Percent Composition • B. The elements contained in ASA are hydrogen, carbon, and oxygen. The combustion of 3.000 g of the pure compound yields 1.200 g of water and 3.72 L of dry carbon dioxide, measure at 750. mmHg and 25 C. Calculate the mass, in g, of each element in the 3.000 g sample. • So primarily, this is a percent composition problem, with the twist that you have the amount of each element produced from the compound’s combustion, not actually in the compound itself.

7. Part B: Percent Composition • The general formula for this reaction looks something like this, unbalanced: • CxHyOz + O2 CO2 + H2O • We know that 1.200 g of water and 3.72 L of CO2 at 750 mmHg and 25 C are produced. Now, let’s think logically. • All of the H in water must have come from the compound. All of the C in CO2 must have also come from the compound. However, the O in both comes from both the compound and oxygen in the air. Thus, when we work backwards (from the products), we can calculate both the amount of C and H in the compound. But because there is excess oxygen, we will subtract the amount of C and H from the total mass of the tablet to find the mass of O in the sample.

8. Part B: Percent Composition • First, convert the mass of water produced to the amount of H in that water. • 1.200 g H2O x (1 mol H2O)/(18.02 g H2O) X (2 mol H)/(1 mol H2O) x (1.01 g H)/(1 mol H) = .1345 g H are in the sample.

9. Part B: Percent Composition • Next, we’ll find the amount of C in the sample by looking at the CO2 produced. • PV = nRT • (750 mmHg / 760 mmHg)(3.72 L) = (n moles CO2)(.08206)(298.15 K) • n = .150 mol CO2 • .150 mol CO2 x (1 mol C)/(1 mol CO2) x (12.01 g C)/(1 mol C) = 1.802 g C are in the sample.

10. Part B: Percent Composition • We have a 3.000 g sample of ASA, made of C, H, and O. • Total mass – mass of H – mass of C = mass of O • 3.000 g - .1345 g – 1.802 g = 1.064 g O • Rounded, there are .135 g H, 1.802 g C, and 1.064 g of O in the sample.

11. Part C: Titration • C. A student dissolved 1.625 g of pure ASA in distilled water and titrated the resulting solution to the equivalence point using 88.43 mL of 0.102 M NaOH(aq). Assuming that ASA has only one ionizable hydrogen, calculate the molar mass of the acid. • Because ASA and NaOH each have one equivalent, a 1:1 ratio of the compounds react. We need to find the moles of OH- that reacted.

12. Part C: Titration • (0.102 M NaOH)(0.08843 L NaOH) = 0.00902 moles NaOH, and therefore there are 0.00902 moles of OH-. • Now, we use our knowledge of the definition of molar mass. Molar mass = (grams of a substance)/(moles of a substance). • We have a 1.635 g sample of ASA. We also know that this sample reacts in a 1:1 ratio with NaOH. Therefore, there are 0.00902 moles ASA in the sample. • Molar mass of ASA = (1.625 g ASA)/(0.00902 moles ASA) = 180. g/mol

13. Part D: More Titrations • D. A 2.00 x 10-3 mole sample of pure ASA was dissolved in 15.00 mL of water and then titrated with 0.100 M NaOH(aq). The equivalence point was reached after 20.00 mL of the NaOH solution had been added. Using the data from the titration, shown in the table below, determine: • (i) the value of the acid dissociation constant, pKa, for ASA • (ii) the pH of the solution after a total volume of 25.00 mL of the NaOH solution had been added (assume that volumes are additive).

14. Part D: More Titrations • i. First things first: we want the pKa of ASA. The pKa is equal to the pH halfway to the equivalence point. • If the equivalence point is reached after 20.00 mL of NaOH have been added, then the pKa when 10.00 mL of NaOH have been added equals 3.44, the pH. • pKa = 3.44 • Ka of ASA = 10-3.44 = 3.6 x 10-4

15. Part D: More Titrations • ii. Beyond the end point, there is excess OH- in the solution. [OH-] determines the pH. 5 mL of OH- were added past the end point. • Moles of excess OH- = (0.005 L)(0.100 M) = 5.00 x 10-4 moles OH-

16. Part D: More Titrations • Next we need to find the overall concentration of OH. • (5.00 x 10-4 moles OH-)/(0.040 L sol) = 1.24 x 10-2M OH-. • pOH = -log(1.24 x 10-2) = 1.90 • pH = 14 – pOH = 12.10 • pH of the solution = 12.10