Chem 2A – Final Exam First letter of your last name NAME: PERM ...

Chem 2A – Final ExamFirst letter of your last nameNAME:___________________________PERM#____________________INSTRUCTIONS: Fill in your name, perm number and first initial of your last nameabove. Be sure to show all of your work for full credit. Use the back of the page ifnecessary.Useful information (periodic table, equations, scratch paper etc..) is on the last threepages.Tear them off for easy reference.1

Chem 2A – Final ExamSCORE: ________ 100 pts possiblePART I – SHORT RESPONSE / MULTIPLE CHOICE1. (2 pts) Consider the following reaction: 2A + B ⎯⎯→ 3C + DIf 3.0 mol of A and 2.0 mol of B react to form 4.0 mol of C. What is the percent yieldof the reaction?a) 50%b) 67%c) 75%d) 89%e) 100%f) none of the above2. (2 pts) What volume of 18.0 M sulfuric acid is required to prepare 16.5 L of 0.126 MH 2 SO 4 ?a) 11.6 mLb) 0.116 Lc) 232 mLd) 0.264 Le) 1.16 Lf) none of the above3. (2 pts) For which acid solution below does 50.0 mL neutralize 50.0 mL of a 0.2 MBa(OH) 2 solution?a) 0.1 M HClb) 0.2 M HNO 3c) 0.3 M HBrd) 0.1 M H 2 SO 3e) 0.2 M H 2 SO 4f) none of the above2

Chem 2A – Final Exam4. (3 pts) What is the oxidation state of chromium in K 2 Cr 2 O 7 ?a) +2b) +3c) +6d) -2e) -3f) none of the above5. (3 pts) According to the equation: 2 KClO3( s) ⎯⎯→ 2 KCl( s) + 3 O2( g)3.00 grams of KClO 3 (122.55 g mol -1 ) decompose and the oxygen at 24.0 o C and0.982 atm is collected. What volume of oxygen gas will be collected, assuming 100%yield?a) 304 mLb) 456 mLc) 608 mLd) 911 mLe) 1820 mLf) none of the above6. (3 pts) 0.05 moles of NaF are dissolved in water to prepare 250 mL of solution. If noother substance is added to the solution which of the following statements is true.a) [H + ] = [OH - ]b) [H + ] = [F - ]c) [Na + ] = [H + ]d) [HF] = [OH - ]e) [OH - ] = [F - ]f) none of the above3

Chem 2A – Final Exam7. (5 pts) Write the molecular formulas for the following compoundsa) manganese (II) hydroxide _________________________b) potassium hydrogen carbonate _________________________c) sodium sulfate _________________________d) ammonium phosphate _________________________e) Ferrous nitrate _________________________8. (5 pts) What is the pH of a 0.2 M solution of sodium acetate, NaC 2 H 3 O 2 ?(for acetic acid - K a = 1.8 × 10 -5 )a) 2.72b) 4.98c) 9.02d) 5.44e) 8.56f) none of the above9. (8 pts) Calculate the concentration of Cu 2+ in a pH = 8.50 buffer solution that has beensaturated with Cu(OH) 22+ -Cu(OH) ( s) Cu ( aq ) + 2 OH ( aq ) K sp = 1.6 × 10 -19a) 1.6 × 10 -2 Mb) 1.8 × 10 -7 Mc) 1.6 × 10 -8 Md) 3.2 × 10 -6 Me) 5.7 × 10 -10 Mf) none of the above24

Chem 2A – Final Exam15. (8 pts) Calculate the pH of a solution that is prepared by adding 19.3 grams of sodiumnitrite (NaNO 2 ) to 500 mL of 0.1 M nitrous acid (HNO 2 , K a = 4.5 × 10 -4 ). Assumethere is no change in volume of solution.16. (7 pts) The sulfide ore of zinc (ZnS) is reduced to elemental zinc by “roasting” it inair to give ZnO followed by heating with carbon monoxide. The two reactionsinvolved are below.3ZnS + O2⎯⎯→ ZnO + SO22ZnO + CO ⎯⎯→ Zn + COIf 5.32 kg of ZnS are treated in this manner and 3.30 kg of pure Zn is obtainedcalculate the percentage yield for this process.28

Chem 2A – Final Exam17. (13 pts) Consider the titration of 50 mL of 0.2 M carbonic acid (H 2 CO 3 ) with 0.5 MNaOH. The titration curve is shown below. K a values are found in the table at the end ofthe exam.a) Label the first and second equivalence points A and B respectively on the plot above.b) Calculate the pH of the solution at the first and second equivalence points.c) Calculate the pH of the solution when 35.0 mL of NaOH have been added to thesolution.9

Chem 2A – Final ExamEquations101,325 Pa = 1 atm760 torr = 1 atmN0= 6.02214×1023−27amu = 1.66054×10 kgPVnTPV=nT1 1 2 21 1 2 2PV= nRTR = 0.08206 L⋅atm⋅mol ⋅KR = 8.314 J ⋅mol ⋅K-1 -1-1 -1Kw = K Kab2⎛ n ⎞⎜P+ a ( V − nb)= nRT2 ⎟⎝ V ⎠nRT nP= −aV − nb V22K = [ H O ][ OH ] = 1.0×10w3+ − −14pH =−log[ H O+ ][ HA]pH = pKa− log [ ]3A −10

Chem 2A – Final ExamAcid Dissociation Constants (25 o C)Name Formula K a1 K a2 K a3Acetic HC 2 H 3 O 2 1.8 × 10 -5Ammonium ion+NH 4 5.6 × 10 -10Arsenic H 3 AsO 4 5.6 × 10 -3 1.0 × 10 -7 3.9 × 10 -12Ascorbic HC 6 H 7 O 6 8.0 × 10 -5 1.6 × 10 -12Benzoic HC 7 H 5 O 2 6.5 × 10 -5Boric H 3 BO 3 5.8 × 10 -10Carbonic H 2 CO 3 4.3 × 10 -7 4.8 × 10 -11Chloroacetic HC 2 H 2 O 2 Cl 1.4 × 10 -3Citric H 3 C 6 H 5 O 7 7.4 × 10 -4 1.7 × 10 -5 4.1 × 10 -7Hydroazoic HN 3 1.9 × 10 -5Hydrocyanic HCN 4.9 × 10 -10Hydrofluoric HF 7.2 × 10 -4Hydrogen selenate ion-HSeO 4 2.2 × 10 -2Hydrogen sulfide H 2 S 9.1 × 10 -8 1.1 × 10 -12Hypobromous HBrO 2.1 × 10 -9Hypochlorous HClO 3.0 × 10 -8Hypoiodous HIO 2.3 × 10 -11Iodic HIO 3 1.7 × 10 -1Lactic HC 3 H 5 O 3 1.4 × 10 -4Malonic H 2 C 3 H 2 O 4 1.5 × 10 -3 2.0 × 10 -6Nitrous HNO 2 4.5 × 10 -4Oxalic H 2 C 2 O 4 5.9 × 10 -2 6.4 × 10 -5Phenol HC 6 H 5 O 1.3 × 10 -10Phosphoric H 3 PO 4 7.5 × 10 -3 6.2 × 10 -8 4.2 × 10 -13Propionic HC 3 H 5 O 2 1.3 × 10 -5Pyrophosphoric H 4 P 2 O 7 3.0 × 10 -2 4.4 × 10 -3Selenous H 2 SeO 3 2.3 × 10 -3 5.3 × 10 -9Sulfuric H 2 SO 4 Strong Acid 1.2 × 10 -2Sulfurous H 2 SO 3 1.7 × 10 -2 6.4 × 10 -8Tartaric H 2 C 4 H 4 O 6 1.0 × 10 -3 4.6 × 10 -5Base Dissociation Constants (25 o C)Name Formula K bAmmonia NH 3 1.8 × 10 -5Aniline C 6 H 5 NH 2 4.3 × 10 -10Dimethlyamine (CH 3 ) 2 NH 5.4 × 10 -4Ethylamine C 2 H 5 NH 2 6.4 × 10 -4Hydrazine H 2 NNH 2 1.3 × 10 -6Hydroxylamine HONH 2 1.1 × 10 -8Methylamine CH 3 NH 2 4.4 × 10 -4Pyridine C 5 H 5 N 1.7 × 10 -9Trimethylamine (CH 3 )CN 6.4 × 10 -5 11

Chem 2A – Final ExamScratch Paper….12