Page%1%of%12%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%multiple9choice%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 A%Page%2%of%12%Thermodynamics:1ΔG°%=%ΔH°%9%TΔS°%ΔH°%=%Σ%ΔH°f%(products)%9%Σ%ΔH°f%(reactants)%ΔS°%=%Σ%S°%(products)%9%Σ%S°%(reactants)%ΔG°%=%Σ%ΔG°f%(products)%9%Σ%ΔG°f%(reactants)%ΔS = qrev/T for%aA%+%bB%%cC%+%dD%%%%%%ΔG%=ΔG°%+%RTln%Q%ΔG°%=%9%RTln%K%ΔG°%=%9%nFΔEº%ΔE%=%%ΔEº%9%(RT/nF)%lnQ%%Ptotal%=%%PA%+%PB%=%XAPA°%+%XBPB°%%Acid1Base:1pH%=%9%log[H3O+]%pX%=%9%log%X%%Quantum:1E%=%hν%λν =%c%Beer’s%Law,%%A%=%εbc%Ideal1Gas:1PV%=%nRT%% Constants:1N0%=%6.02214%×%1023%mol91%Ry%=%2.179874%×%10918%J%=%1312%kJ%mol91%Ry%=%3.28984%×%1015%Hz%k%=%1.38066%×%10923%J%K91%h%=%6.62608%×%10934%J%s%me%=%9.101939%×%10931%kg%c%=%2.99792%×%108%m%s91%1eV%=%1.602×10919%J%1%J%=%1%kg%m2%s92%T%(K)%=%T%(C)%+%273.15%F%=%96,485%C%/%mol%1%V%=%1%J%/%C%1%nm%=%1099%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%K91%mol91%R%=%8.20578%×%1092%L%atm%K91%mol91%%%800%600%400%200%IR%Red%Green%Blue%UV%Wavelength%(nm)%Color and Wavelength of LightPage%3%of%12%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%12%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− %%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% %Name_________________________________GSI_____________ Page%5%of%12%CALCIUM PHOSPHATE SOLUBILITY Teeth are made mainly of hydroxyapatite, Ca5(PO4)3(OH). Consider the solubility of hydroxyapatite. Ca5(PO4)3(OH) (s)  5Ca2+ (aq) + 3PO43− (aq) + OH− (aq) Ksp ~ 10-36 11. Which equation can you use to determine the solubility, x, of Ca5(PO4)3(OH)? A) Ksp = 15x B) Ksp = x9 C) Ksp = 15x9 D) Ksp = (5)5(3)3x9 SHORT ANSWER: Bacteria in your mouth decompose sugars to form lactic acid,