CHEM 121b Exam 4 Spring 1999
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1 Name SSN CHEM 121b Exam 4 Spring 1999 This exam consists of 10 multiple choice questions (each worth 2 points), and 6 written problems (points noted below). There are a total of 100 possible points. Carefully mark the appropriate answer to each question on the Scantron sheet, and show any work in the space provided. Please hand in the Exam, the Scantron sheet, and your 3.5 x 5 card. Multiple Choice 1. Identify the strongest acid. a. HCN. c. H 2 O. b. HNO 3. d. OH. 2. Which of the following is true for the dissociation of a weak acid? a. K a is large. c. the equilibrium lies far to the left. b. the equilibrium lies far to the right. d. [H + ] >> [HA]. 3. The conjugate acid and conjugate base of bicarbonate ion, HCO 3, are, respectively: a. H 3 O + and OH. c. H 2 CO 3 and OH. b. H 3 O + and CO 2 3. d. H 2 CO 3 and CO Which factor listed below is most important in determining the strength of an oxyacid? a. the size of the molecule. b. the ability of the molecule to change atomic orientation. c. the identity of the central atom in the molecule. d. the number of oxygen atoms present in the molecule. 5. If sodium nitrate, NaNO 3, is dissolved in water, the resulting solution is: a. acidic. c. basic. b. neutral. d. not enough information to tell. 6. If ammonium perchlorate, NH 4 ClO 4, is dissolved in water, the resulting solution is: a. acidic. c. basic. b. neutral. d. not enough information to tell. 7. Which of the following is true for a buffered solution? a. the solution resists change in its [H + ]. b. the ph will change drastically if a small amount of acid is added. c. the ph will change drastically if a small amount of base is added. d. the solution consists of a strong acid and the salt of its conjugate base. 8. If 25.0 ml of M NaOH solution is titrated with M H 2 SO 4, the principal components in the flask (besides the indicator phenolphthalein) at the endpoint will be: a. sodium hydroxide, sulfuric acid, and water. b. sodium sulfate and water. c. sodium hydroxide, sodium sulfate, and water. d. sodium sulfate, sulfuric acid, and water.
2 Chem 121b Exam 4 Page 2 9. Consider the titration of equal volumes of 0.1 M HCl and 0.1 M HC 2 H 3 O 2 with 0.1 M NaOH. Which of the following would be the same for both titrations? a. the initial ph. b. the ph at the halfway point. c. the ph at the equivalence point. d. the volume of NaOH added to reach the equivalence point. 10. Methyl orange is an indicator with a K a of 1.0 x Its acid form, HIn, is red, while its base form, In, is yellow. At ph 6.0, the indicator will be: a. red. c. yellow. b. orange. d. blue. Written Problems Write out the solution to each written problem. Please show all of your work. 11. Calculate the [H 2 NNH 2 ], [H 2 NNH 3 + ], [OH ], [H + ], poh, and ph for a 0.15 M solution of the weak base hydrazine. (15 pts.) [H 2 NNH 2 ] = [H 2 NNH 3 + ] = [OH ] = [H + ] = poh = ph =
3 Chem 121b Exam 4 Page Calculate the [H 2 CO 3 ], [HCO 3 ], [CO 3 2 ], [H + ], [OH ], and ph for a 0.25 M solution of carbonic acid. (15 pts.) [H 2 CO 3 ] = [HCO 3 ] = [CO 3 2 ] = [H + ] = [OH ] = ph =
4 Chem 121b Exam 4 Page Calculate the [HClO], [ClO ], [Na + ], [H + ], [OH ], and ph for a buffer solution containing 0.25 M hypochlorous acid and 0.35 M sodium hypochlorite. (15 pts.) [HClO] = [ClO ] = [Na + ] = [H + ] = [OH ] = ph =
5 Chem 121b Exam 4 Page Calculate the ratio [C 6 H 5 NH 2 ]/[C 6 H 5 NH 3 + ] needed to make a ph 5.00 buffered solution from aniline. (10 pts.) 15. The salt Ag 2 CrO 4 has a K sp = 9.0 x at 25 C. What is its solubility at this temperature? (10 pts.)
6 Chem 121b Exam 4 Page Consider the titration of 50.0 ml of M HCl with M NaOH. (HCl is in the Erlenmeyer flask and NaOH is in the buret.) A. What is the ph of the HCl solution before any NaOH is added? (3 pts) B. What is the ph after 20.0 ml of NaOH is added? (5 pts) C. How many ml of NaOH must be added to reach the equivalence point? (2 pts) D. What is the ph after 35.0 ml of NaOH is added? (5 pts) Multiple Choice (20) Problem 11 (15) Problem 12 (15) Problem 13 (15) Problem 14 (10) Problem 15 (10) Problem 16 (15) Exam 4 Total (100)
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FORMULA SHEET (tear off) N A = 6.022 x 10 23 C = ( 5 / 9) ( F - 32) F = ( 9 / 5)( C) + 32 1 amu = 1.661 x 10-27 kg C = K - 273.15 K = C + 273.15 1 atm = 760 torr = 760 mm Hg 1 atm = 1.013 bar pv = nrt
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