Page 1 of 14 Chemistry 1A, Fall 2012 Final Exam December 10, 2012 (180 min, closed book) Name:__________________________________ SID:___________________________________ GSI Name:________________ Disc. Day/Time:_______ • The test consists of short answer questions and 40 multiple-choice questions. • Put your written answers in the boxes provided. Answers outside the boxes may not be considered in grading. ONLY THE FIRST 20-30 WORDS WILL BE GRADED • Write your name on every page of the exam. Question Page Points Score Multiple Choice (1-40) 3-12 160 Acid-base equilibrium 4 4 Solubility 5 8 Bond energy 6 4 Free energy 7 4 Potential energy 8 6 Molecular orbitals 10 4 Electrochemical cells 11 12 Decomposition equilibrium 12 8 Total 210 APage 2 of 14 Thermodynamics: ΔG° = ΔH° - TΔS° ΔH° = Σ ΔH°f (products) - Σ ΔH°f (reactants) ΔS° = Σ S° (products) - Σ S° (reactants) ΔG° = Σ ΔG°f (products) - Σ ΔG°f (reactants) ΔS = qrev/T for aA + bB ⇌ cC + dD ΔG =ΔG° + RTln Q ΔG° = - RTln K ΔG° = - nFΔEº ΔE = ΔEº - (RT/nF) lnQ Ptotal = PA + PB = XAPA° + XBPB° Acid Base: pH = - log[H3O+] pX = - log X badcBADCQ][][][][=RSTRHK°Δ+°Δ−=1ln][][logHAApKpHa−+=Quantum: E = hν λν = c Beer’s Law, A = εbc Ideal Gas: PV = nRT RTEkin23=M3RTvrms=Constants: N0 = 6.02214 × 1023 mol-1 Ry = 2.179874 × 10-18 J = 1312 kJ mol-1 Ry = 3.28984 × 1015 Hz k = 1.38066 × 10-23 J K-1 h = 6.62608 × 10-34 J s me = 9.101939 × 10-31 kg c = 2.99792 × 108 m s-1 1eV = 1.602×10-19 J 1 J = 1 kg m2 s-2 T (K) = T (C) + 273.15 F = 96,485 C / mol 1 V = 1 J / C 1 nm = 10-9 m 1 kJ = 1000 J 1 atm = 760 mm Hg = 760 torr ≈ 1 bar 1 L atm ≈ 100 J Gas Constant: R = 8.31451 J K-1 mol-1 R = 8.20578 × 10-2 L atm K-1 mol-1 800 600 400 200 IR Red Green Blue UV Wavelength (nm) Color and Wavelength of LightName_________________________________GSI_____________ Page 3 of 14 OHPOHHO OOHPOHHO OPHOSPHORIC ACID STRUCTURE Two structures for phosphoric acid, H3PO4 are shown here. Both are structures that you can find in reference books to describe this molecule. Note: You will need to add lone pairs on the O atoms. 1. Which statement is correct for the structure on the left? A) It satisfies the octet rule, and there are no formal charges. B) It does not satisfy the octet rule, and there are no formal charges. C) It satisfies the octet rule, and there is a formal charge of +1 on the P atom and a formal charge of -1 on one O atom. D) It does not satisfy the octet rule, and there is a formal charge of +1 on the P atom and a formal charge of -1 on one O atom. 2. Which statement is correct for the structure on the right? A) It satisfies the octet rule, and there are no formal charges. B) It does not satisfy the octet rule, and there are no formal charges. C) It satisfies the octet rule, and there is a formal charge of +1 on the P atom and a formal charge of -1 on one O atom. D) It does not satisfy the octet rule, and there is a formal charge of +1 on the P atom and a formal charge of -1 on one O atom. 3. The average phosphorus-oxygen bond length in H3PO4 is less than a P−O single bond. Many textbooks make the claim that the structure on the right with P=O is the best representation for H3PO4. Do you agree with this claim? A) I agree. P=O double bonds are shorter than P-O single bonds. B) I do not agree. P=O double bonds are longer than P-O single bonds. C) I do not agree. Hydrogen bonds cause the phosphorus-oxygen bonds to be shorter in both representations of the molecule. D) I think that both are good models. The formal charges on the P and O atoms for the structure on the left can also explain a shorter average bond length. 4. A concentrated aqueous solution of phosphoric acid is extremely viscous. It pours like syrup. How can you explain this observation? A) There are strong hydrogen bonds with water, which restrict molecular motion. B) Phosphoric acid is a polymer. The polymer chains get tangled with one another. C) Phosphoric acid is a strong acid and therefore, completely dissociated into ions. D) There are strong London dispersion forces because the molecule has many electrons.Name_________________________________GSI_____________ Page 4 of 14 PHOSPHORIC ACID EQUILIBRIA H3PO4 (aq) + H2O (l) ! H+ (aq) + H2PO4− (aq) Ka1 = 7.3×10−3 H2PO4− (aq) + H2O (l) ! H+ (aq) + HPO42− (aq) Ka2 = 6.3×10−8 HPO42− (aq) + H2O (l) ! H+ (aq) + PO43− (aq) Ka3 = 4.0×10−13 5. What is the pH of an aqueous solution of phosphoric acid, H3PO4? A) pH = -log Ka1 = 2.14 B) The pH is about 1 because phosphoric acid is a strong acid. C) The pH depends on the concentration. D) The H+ concentration is 3 times greater than in a HCl solution of equal concentration. 6. What is the pH of a solution that is 0.010 M H3PO4 and 0.010 M Na3PO4? A) 2 B) 2.14 C) 7.2 D) 12.4 7. Phosphoric acid is used to give cola its tangy flavor. The pH of cola is 3.0. What is the concentration of phosphoric acid? A) 3.0 M B) 1.0 × 10−3 M C) 7.3 × 10−3 M D) 1.4 × 10−4 M 8. The pH of 100 mL of a 0.10 M H3PO4 solution is being adjusted to pH = 7 by adding NaOH (s). How much NaOH (s) must be added? A) 0.006 mol B) 0.014 mol C) 0.020 mol D) 0.030 mol 9. What color is a dilute solution of bromothymol blue indicator in an aqueous solution of 0.0010 M H3PO4? Ka(bromothymol blue) = 1.0 ×10-7. The acid form of the indicator is yellow, and the base form is blue. A) yellow B) blue C) green 10. Which species is present in the highest concentration in a solution of Na3PO4 at pH = 14? A) H3PO4 B) H2PO4− C) HPO42− D) PO43−Name_________________________________GSI_____________ Page 5 of 14 SHORT ANSWER: A can of soda is 10−4 M H3PO4 and 10−1 M CO2. When cola goes flat, the CO2 concentration decreases from 10−1 M to 10−2 M, and one might expect a change in pH. However, the pH does not change. Use the data provided below to explain why. H3PO4 (aq) + H2O (l) ! H+ (aq) + H2PO4− (aq) Ka1 = 7.3×10−3 CO2 (aq) + H2O (l) ! H+ (aq) + HCO3− (aq) Ka1 = 4.5×10−7 The total H+ concentration of this solution is determined by H3PO4 because the CO2 equilibrium is shifted almost completely to the left. Regardless of its concentration, CO2 would not contribute to the total H+. (2pts) recognizing H3PO4 as the stronger acid (2pts) recognizing that CO2, does not contribute because …