Answers & Clues to Final Exam Revision Questions: 1. Arrange the following in order of increasing boiling point: RbCl, CH3Cl, CH3OH, CH4. Ans: CH4 < CH3Cl < CH3OH < RbCl CH4 – non polar: dispersion force only CH3Cl – polar: dispersion & dipole-dipole forces CH3OH – polar: dispersion, dipole-dipole & H-bonding RbCl – ionic salt that contains ionic bonding 2. Which one of the following substances should exhibit hydrogen bonding in the liquid state? SiH4 H2 H2S CH4 CH3NH2 Ans: CH3NH2 3. Calculate the molality of a solution containing 14.3 g of NaCl in 42.2 g of water. Ans: ๐๐๐๐๐๐๐ก๐ฆ = 4. 14.3 ๐ ⁄58.5 ๐/๐๐๐ 0.0422 ๐๐ ๐๐๐๐ ๐ ๐๐๐ข๐ก๐ = ๐๐๐ ๐ ๐ ๐๐๐ฃ๐๐๐ก (๐๐) = 5.80 ๐ Calculate the approximate freezing point of a solution made from 21.0 g NaCl and 1.00 ๏ด 102 g of H2O. Kf of water is 1.86๏ฐC/m. Ans: ๐๐๐๐๐๐๐ก๐ฆ = 21.0 ๐ ⁄58.5 ๐/๐๐๐ 0.100 ๐๐ NaCl ๏ Na+ + Cl– = 3.5897 ๐ van’t Hoff factor, i = 2 โTf = (i)(Kf)(m) = (2)( 1.86๏ฐC/m)(3.5897 m) = 13.354 oC Freezing point of water = 0 oC ๏Freezing point of solution = 0 oC – 13.354 oC = –13.354 oC 5. A first-order reaction has a rate constant of 3.00 ๏ด 10๏ญ3 s๏ญ1. The time required for the reaction to be 75.0% complete is Ans: if 75% complete: [A]0 = 100% and [A]t = 25% ln [A]t = ln [A]0 – kt ln 25 = ln 100 – (3.00 ๏ด 10๏ญ3 s๏ญ1) (t) ๏t = 462.09 s 6. When the concentrations of reactant molecules are increased, the rate of reaction increases. Explain why. Ans: As the reactant concentration increases, the frequency of molecular collisions increases. 7. If one starts with pure NO2(g) at a pressure of 0.500 atm, the total pressure inside the reaction vessel when 2NO2(g) 2NO(g) + O2(g) reaches equilibrium is 0.674 atm. Calculate the equilibrium partial pressure of NO2. Ans: 2NO2(g) 2NO(g) + O2(g) I 0.500 atm 0 0 C –2x +2x +x E 0.500 – 2x 2x x At equilibrium, total pressure = 0.674 atm P(NO2) + P(NO) + P(O2) = 0.674 atm (0.500 – 2x) + 2x + x = 0.674 atm ๏x = 0.674 atm – 0.500 atm = 0.174 atm 8. At 340 K, Kp = 69 for the reaction H2 (g) + I2 (g) 2HI (g). 50.0 g of HI is injected into an evacuated 5.00-L rigid cylinder at 340 K. What is the total pressure inside the cylinder when the system comes to equilibrium? Ans: Calculate initial pressure of HI: ๐ = Given: H2 (g) + I2 (g) 2HI (g) ๐๐ ๐ ๐ = ( 50.0 ๐๐ก๐ )(0.0821 ๐ฟ. .๐พ)(340๐พ) 127.908 ๐๐๐ 5.00 ๐ฟ = 2.182 ๐๐ก๐ Kp = 69 HI is injected first, therefore HI now becomes the reactant, the equation has to be inversed: 2HI (g) H2 (g) + I2 (g) 2HI (g) ๐พ๐ = Kp’ = 1/69 = 0.0145 H2 (g) + I2 (g) I 2.182 atm 0 0 C –2x +x +x E 2.182 – 2x x x ๐๐ป2 ×๐๐ผ2 ๐ฅ 2.182−2๐ฅ 2 ๐๐ป๐ผ = (๐ฅ)(๐ฅ) (2.182−2๐ฅ)2 = √0.0145 = 0.0145 ๏x = 0.212 ๏Total pressure = (2.182 – 2(0.212)) + 0.212 + 0.212 = 2.18 atm 9. Acid strength decreases in the series HI > HSO4๏ญ > HF > HCN. Which of the following anions is the weakest base? Ans: I๏ญ Clue: Use conjugate pair: the weakest base is the anion that comes from the strongest acid. 10. Arrange the acids HOCl, HClO3, and HClO2 in order of increasing acid strength. Ans: HOCl < HClO2 < HClO3 Clue: apply the concept of increasing oxidation number, refer to chlorine – higher oxidation number means stronger acid. 11. Which of the following yields a basic solution when dissolved in water? KNO2 NH4Cl NaCl SO2 Ans: KNO2 Clue: write the hydrolysis equations and find out which ion (H+ or OH–) exist: KNO2 ๏ K+ + NO2– K+ come from strong base (KOH), it will not hydrolyzed. NO2– + H2O โ HNO2 + OH– (OH– is present, ๏solution is basic) 12. Calculate the concentration of oxalate ion (C2O42๏ญ) in a 0.175 M solution of oxalic acid (C2H2O4). [For oxalic acid, Ka1 = 6.5 ๏ด 10๏ญ2, Ka2 = 6.1 ๏ด 10–5.] Ans: ๐พ๐1 = โ H2C2O4 HC2O4– H+ I 0.175 M 0 0 C –x +x +x E 0.175 – x x x [๐ป + ][๐ป๐ถ2 ๐4− ] [๐ป2 ๐ถ2 ๐4 ] = ๐ฅ2 0.175−๐ฅ = 6.5 × 10−2 *Clue: solve x using quadratic equation โ HC2O4– ๐พ๐2 = + C2O42– ๏x = 0.078995 or –0.143995 + H+ I 0.078995 M 0 0.078995 M C –x +x +x E 0.078995 – x x 0.078995 + x [๐ป + ][๐ถ2 ๐42− ] [๐ป๐ถ2 ๐4− ] = (๐ฅ)(0.078995+๐ฅ) 0.078995−๐ฅ = 6.1 × 10−5 *Clue: solve x using assumption. Check validity! ๏x = [C2O42–] = 6.1 ๏ด 10–5 13. Consider the weak bases below and their Kb values: C6H7O Kb = 1.3 ๏ด 10๏ญ10 C2H5NH2 Kb = 5.6 ๏ด 10๏ญ4 C5H5N Kb = 1.7 ๏ด 10๏ญ9 Arrange the conjugate acids of these weak bases in order of increasing acid strength. Ans: C2H5NH3+ < C5H5NH+ < C6H7OH Clue: arrange the weak bases in order of increasing basic strength (higher Kb value means stronger base): C6H7O < C5H5N < C2H5NH2 Apply this concept: ‘Strong base has weak conjugate acid pair.’ Therefore, the strongest base has the weakest conjugate acid. 14. When 2.0 ๏ด 10–2 mole of nicotinic acid (a monoprotic acid) is dissolved in 350 mL of water, the pH is 3.05. What is the Ka of nicotinic acid? Ans: molarity of nicotinic acid = 2.0 ๏ด 10–2 mol/0.350 L = 0.057 M Assume nicotinic acid = HA HA I 0.057 M C –x E 0.057 – x pH = 3.05 ๐พ๐ = 15. โ (๐ฅ)(๐ฅ) A– + 0 0 +x +x x x ; –log [H+] = 3.05 = 0.057−๐ฅ H+ (8.913 × 10−4 )2 0.057− (8.913 × 10−4 ) therefore [H+] = 8.913 ๏ด 10–4 M = x = 1.42 × 10−2 Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of cyanic acid (HCNO) and 0.80 mole of sodium cyanate (NaCNO) in enough water to make 1.0 liter of solution. [Ka(HCNO) = 2.0 ๏ด 10๏ญ4] [๐ถ๐๐ − ] 0.80 Ans: ๐๐ป = ๐๐พ๐ + log [๐ป๐ถ๐๐] = 3.70 + log 0.20 = 4.30 16. Consider a buffer solution prepared from HOCl and NaOCl. Which is the net ionic equation for the reaction that occurs when NaOH is added to this buffer? Ans: OH๏ญ + HOCl โ H2O + OCl๏ญ Clue: OH– from the NaOH added reacts with the acid in the buffer solution 17. You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH3COOH) and 0.20 M sodium acetate (CH3COONa). What will the pH of this solution be after the addition of 20.0 mL of 1.00 M NaOH solution? [Ka = 1.8 ๏ด 10๏ญ5] Ans: Mole CH3COOH = Mole CH3COO– = I C E 1000 0.20 ×500 1000 1.00 ×20 Mole NaOH = CH3COOH 0.150 mol –0.020 mol 0.130 mol 0.30 ×500 OH– โ 0.020 mol –0.020 mol 0 + Molarity CH3COO– = = 0.100 ๐๐๐ = 0.020 ๐๐๐ 1000 Molarity CH3COOH = = 0.150 ๐๐๐ 0.130 ๐๐๐ 0.520 ๐ฟ 0.120 ๐๐๐ 0.520 ๐ฟ CH3COO– 0.100 mol +0.020 mol 0.120 mol + H2O = 0.250 ๐ = 0.231 ๐ [๐ถ๐ป3 ๐ถ๐๐− ] 0.231 ๐๐ป = ๐๐พ๐ + log = 4.74 + log = 4.71 [๐ถ๐ป3 ๐ถ๐๐๐ป] 0.250 18. Calculate the pH at the equivalence point for the titration of 0.20 M HCl with 0.20 M NH3 (Kb = 1.8 ๏ด 10๏ญ5). Ans: assume 1 L of each solution: HCl + NH3 โ NH4Cl I 0.20 mol 0.20 mol 0 C –0.020 mol –0.020 mol +0.020 mol E 0 0 0.20 mol The salt NH4Cl will undergo hydrolysis: Molarity of NH4Cl = 0.20 mol / 2 L = 0.10 M NH4+ + 0.10 M –x 0.10 – x I C E ๐พ๐ = H2O ๐พ๐ค 1.0 × 10−14 = ๐พ๐ 1.8 × 10−5 โ NH3 + H3O+ 0 0 +x +x x x (๐ฅ)(๐ฅ) = 0.10 − ๐ฅ ๐ฅ2 ∴ 0.10−๐ฅ = 5.56 × 10−10 ๐ฅ = [๐ป3 ๐+ ] = 7.45 × 10−6 pH = –log 7.45 ๏ด 10–6 = 5.13 19. The molar solubility of magnesium carbonate is 1.8 ๏ด 10๏ญ4 mol/L. What is Ksp for this compound? Ans: 20. MgCO3 โ Mg2+ + CO3–2 s s Ksp = [Mg2+][ CO3–2] = s2 = (1.8 ๏ด 10๏ญ4)2 = 3.24 ๏ด 10–8 Solid sodium iodide is slowly added to a solution that is 0.0050 M Pb2+ and 0.0050 M Ag+. (Ksp (PbI2) = 1.4 ๏ด 10๏ญ8; Ksp (AgI) = 8.3 ๏ด 10๏ญ17). What compound will precipitate first? Ans: AgI will precipitate first because AgI has the smallest Ksp. 21. Calculate the minimum concentration of Mg2+ that must be added to 0.10 M NaF in order to initiate a precipitate of magnesium fluoride. (For MgF2 , Ksp = 6.9 ๏ด 10๏ญ9.) Ans: MgF2 โ Mg2+ + 2F– s 2s 2+ 2 – Ksp = [Mg ][F ] = 6.9 ๏ด 10๏ญ9 [Mg2+](0.10)2 = 6.9 ๏ด 10๏ญ9 ๏ [Mg2+] = 6.9 ๏ด 10๏ญ7 22. Will a 0.1 M solution of Na2HPO4 (aq) be acidic, basic or neutral? Given: Ka = 4.8 ๏ด 10–13 and Kb = 1.6 ๏ด 10–7. Ans: ๏ท Na+ is a spectator ion (it comes from strong base, NaOH), so it does not affect the acidity or basicity of a solution. ๏ท There are two possible reactions HPO4– ion can undergo: ๏ท 23. #1: HPO4–2 โ H+ + PO4–3 Ka = 4.8 ๏ด 10–13 #2: HPO4–2 + H2O โ H2PO4– + OH– Kb = 1.6 ๏ด 10–7 Kb > Ka, therefore the solution is basic. What is the missing symbol in this plutonium fission reaction? 239 1 91 1 ๏ฎ Pu + n ______ + Sr + 3 94 0 38 0n 146 Ans: 56 Ba 24. Radium-226 decays by alpha emission. What is its decay product ? 222 Ans: 86 Rn 25. Find the nuclear binding energy of potassium-40 (atomic mass = 39.9632591 amu) in units of joules per nucleon. Data: neutron mass = 1.674928 ๏ด 10–24 g; proton mass = 1.672623 ๏ด 10–24 g; electron mass = 9.109387 ๏ด 10–28 g; NA = 6.0221367 ๏ด 1023 /mol; c = 2.99792458 ๏ด 108 m/s; 1 g = 6.022 ๏ด 1023 amu; 1 J = 1 kg/m2.s2. Ans: 40 19๐พ โm = [(19 ๏ด 1.672623 ๏ด 10–27 kg) + (21 ๏ด 1.674928 ๏ด 10–27 kg)] 1๐ – (39.9632591 amu ๏ด 6.022 × 1023๐๐๐ข × → 19 11๐ + 21 10๐ 1 ๐๐ 1000 ๐ ) = 5.912208818 ๏ด 10–28 kg โE = โm ๏ด c2 = 5.912208818 ๏ด 10–28 kg ๏ด (2.99792458 ๏ด 108 m/s)2 = 5.3136 ๏ด 10–11 J Binding energy per nucleon: โE / 40 = 5.3136 ๏ด 10–11 /40 = 1.328 ๏ด 10–12 J per nucleon 26. A nucleus of carbon-14 lies above the belt of stability. Write an equation for radioactive decay of carbon-14. Ans: 146๐ถ → 27. 14 7๐ + 0 −1๐ฝ A radioactive isotope of Rb-87 decays as follow: (cancel this sentence) 25 11๐๐ → 25 12๐๐ + 0 −1๐ฝ t½ = 15 hours Starting with 100 g of Na-25, calculate the quantity of Mg-25 produced after 19 hours. Ans: Initial mass of Na-25 = 100 g At present, assume mass of Mg-25 = x g, therefore mass of Na-25 = (100 – x) g λ = ln 2 / 15 hours = 0.0462 hours–1 Calculating for Na-25: ln Nt = ln N0 – λt ln (100 – x) = ln 100 – (0.0462 hours–1)(19 hours) ln (100 – x) = 2.7274 (100 – x) = e 2.7274 ๏x = 41.57 g = mass Mg-25 Mass Na-25 left = 100 g – 41.57 g = 58.43 g 28. What is the monomer of natural rubber? Draw its structure and name it. Ans: monomer of natural rubber is isoprene. 2-methylbuta-1,3-diene 29. K2Cr2O7 / H2SO4 Reflux CH2ClCH2–OH + 2[O] ? ? + H2O Ans: CH2ClCOOH (chloroethanoic acid) 30. ๐๐ CH2 = CH– CH3 + H2 → ? Ans: CH3 – CHBr – CH3 (2-bromopropane) 31. Ans: 32. (cyclopropene) Write chemical equations to show the steps involve in the mechanism for the polymerization of polyethylene using benzoyl peroxide radical as the initiator. Ans: Step 1: Initiation Or can be written as: C6H5COO–OOCC6H5 ๏ง 2(C6H5COO•) Reaction of radical with monomer (ethylene): C6H5COO• + CH2 = CH2 ๏ง C6H5COO– CH2– CH2• Step 2: Propagation C6H5COO– CH2– CH2• + CH2 = CH2 ๏ง C6H5COO– CH2– CH2– CH2– CH2• Step 3: Termination C6H5COO– CH2– CH2– CH2 n – CH2• + •CH2–CH2– CH2– CH2 ๏ง C6H5COO– CH2– CH2– CH2 n – CH2 –CH2–CH2– CH2– CH2 m – OOCC6H5 m – OOCC6H5 From nuclear tutorial: 24.43 A rock contains 270 ๏ญmol 238U ( t1 / 2 = 4.5 ๏ด 109 years) and 110 ๏ญmol 206Pb. Assuming that all the 206Pb comes from the decay of the 238U, estimate the age of the rock. Ans: Initial % of 238U = 100% 270๐๐๐๐ % of 238U at present = (270+110)๐๐๐๐ × 100% = 71.05% λ = ln 2 / 4.5 ๏ด 109 years = 1.54 ๏ด 10–10 yrs–1 ln Nt = ln N0 – λt 100 71.05 ∴๐ก= = 2.22 × 109 ๐ฆ๐๐๐๐ 1.54 × 10−10 ln