EXAM 2 CHEMISTRY 224 March 1, Use a #2 pencil to code all information on the answer sheet.

Size: px

Start display at page:

Download "EXAM 2 CHEMISTRY 224 March 1, Use a #2 pencil to code all information on the answer sheet."

Herbert Gilmore
5 years ago
Views:

1 1. Read the following instructions carefully EXAM CHEMISTRY March 1, 01. Write your name and Purdue ID number on the answer sheet 3. Write your Graduate Instructor s name on the line for Instructor on the answer sheet. Use a # pencil to code all information on the answer sheet. 5. Code your name and Purdue ID number on the answer sheet 6. Code 0001 as the Section number for Tue. lab section, 000 for Wed. lab section, 0003 for Thurs. lab section. 7. Code the one best answer to each question on the answer sheet. THERE ARE 0 QUESTIONS 1 BONUS QUESTION. EACH QUESTION IS WORTH TEN (10) POINTS YOU ARE SUPPOSED TO KNOW (REMEMBER) A NUMBER OF BASIC EQUATIONS. ADDITIONAL FORMULAS AND REFERENCE TABLES ARE INCLUDED AT THE END OF THE EXAM MANY PROBLEMS CAN BE SOLVED ONLY USING THE REFERENCE MATERIALS IN EACH CASE IT IS YOUR RESPONSIBILITY TO CONSULT THE REFERENCE SECTION AT THE END OF THE EXAM. IF IT SEEMS THAT YOU NEED A CERTAIN COMPLICATED FORMULA WHICH IS NOT GIVEN IN THE TEXT OF THE EXAM WE SUGGEST THAT YOU HAVE A CLOSER LOOK AT THE PROBLEM PERHAPS, IT CAN BE SOLVED IN A SIMPLER WAY.

2 1. The concentration of H in a solution is M. What is the ph of this solution? a. 6. b. 3. c.. d..0 e. 10. Answer: b. ph=-log([h ]).. Calculate the approximate ph of a buffer solution made from 0.5 M acetic acid and 0.50 M sodium acetate. a..5 b..5 c..8 d. 5.1 e. 7.5 Answer: d. ph = pka log ([NaAc]/[HAc]) (Henderson-Hasselbalch) ph = -log(1.75e-5) log(0.50/0.5) (see Appendix 3 for Ka value) ph = A solution is prepared by combining 0.5 mm of NaOH and 0.55 mm of HCl. What is the concentration [H ] in this solution? a mm b mm c mm d M e. 0.5 mm Answer: c. The excess of HCl over NaOH. Autoprotolysis can be neglected (it produces much less than 0.05 mm of H ).. Five different acids were used to prepare 0.1M solutions. Which of these solutions is most acidic? a. Propanoic acid b. Trimethyl ammonium ion c. Formic acid d. Hydrogen sulfide e. All above solutions are equally acidic

3 Answer: c. These are all weak acids where [ H ] = Kc a HA, hence we should just pick the acid with highest K a. We could also say that right away, since the acid with higher K a dissociates more fully. See Appendix 3 for K a values. 5. Use the picture below to determine the wavelength of the ocean wave: a. approx. 0.3 m b. approx. 1 m c. approx. m d. approx. 7.6 m e. approx. 19 m Answer: b. about wave ridges over the length of 19-0 m 6. Please place different types of electromagnetic waves in the order of longest to shortest wavelengths: a. radio waves, visible rays, infrared rays, ultraviolet rays, x-rays, b. radio waves, x-rays, visible rays, ultraviolet rays, infrared rays c. x-rays, visible rays, ultraviolet rays, infrared rays, radio waves d. x-rays, ultraviolet rays, visible rays, infrared rays, radio waves e. radio waves, infrared rays, visible rays, ultraviolet rays, x-rays Answer: e. See the Electromagnetic Spectrum diagram at the end of the exam. 7. Find the concentration of three metal solutions by using the information below (similar to what you did in lab 6): A λ = β λ,ni C Ni β λ,co C Co β λ,cu C Cu α where α = 0.05, β Ni = 0.1 M -1, β Co = 0. M -1, β Cu = 0.15 M -1

4 The measured absorbance was 0.8 (wavelength 67 nm, path length cm). a. Ni = 5.M, Co = 0.5M, Cu = 0.9M b. Ni =.0M, Co = 0.5M, Cu = 5.M c. Ni =.6M, Co = 0.M, Cu = 0.50M d. Ni = 10.8M, Co = 0.9M, Cu = 1.98M e. None of above Answer: a. Plug in the numbers. 8. Which molecule(s) show amphiprotic characteristic in solution? Please provide most complete answer. a. H O b. HSO c. NH CH COOH d. both a and b e. all of above Answer: e. H O has been discussed in class. HSO was also discussed in class and also reminiscent of lab 5 (titration of diprotic acid). NH CH COOH can turn into NH3CHCOOH (recall ammonium ion), as well as NHCHCOO (recall phenolphthalein, etc). 9. The reaction HSO HO H3O HSO in dilute aqueous solution is characterized by the equilibrium constant K eq = 18. Based on this information, please determine the acid dissociation constant K a1 for sulfuric acid. a. 1M b. 10 M c. 100 M d M e M -1 Answer: d. the standard equilibrium constant is: K eq [ H3O ][ HSO] = [ H SO ][ H O] whereas the acid dissociation is:

5 K a1 [ H3O ][ HSO] = [ H SO ] Hence simply multiply K eq =18 by [H O], where [H O]=1000 (g/l)/18(g/mol) 10. Using a ph meter, the reading for a 500 ml buffer solution of NaHCO 3 and Na CO 3 was What is the mass of NaHCO 3 added to the buffer solution to obtain this measured ph, given that the concentration of Na CO 3 in this solution is 0.1 M? Assume: For H CO 3 Ka 1 = 5x10-7 M Ka = 5x10-11 M MW H CO 3 = 6 g/mol MW Na = 3 g/mol a. 5.3 grams b. 3.9 grams c. 1.7 grams d. 0. grams e. None of the above Answer: e. Henderson-Haselbalch equation for the equilibrium utilizing Ka reads = log(5 10 ) log( ) [NaHCO 3] From here [Na CO 3 ]= M. Using MW of NaHCO 3, 6-13 =8 g/mol, and considering that we need to prepare only 500 ml of the buffer, obtain.65 g. 11. Analyst Alice needs a phosphate buffer with ph=6.8. She prepares the buffer according to the following recipe: PBS stock solution [0 x]: 0. M Phosphate Buffered Saline (PBS) Dissolve 3 g Na HPO H 0 6 g NaH PO H 0 16 g NaC1 in 900 ml distilled water (intensive stirring and some heating will speed dissolution). After the solution is prepared, Alice measures the ph and obtains ph=7.. What is the best course of action? a. Discard the solution and carefully prepare a new one b. Select a different type of buffer to ensure ph=6.8 c. Adjust ph by adding a small amount of concentrated NaOH d. Adjust ph by adding a small amount of concentrated HCl e. Adjust ph by adding a small amount of NaCl

6 Answer: d. It is ok to use buffer near the intended ph target. 1. What happens to [H ] of pure water as the temperature is raised from 5 o C to 55 o C? a. [H ] as chemical bonds in H O are weakened at high temperature b. [H ] since association between H and OH - increases c. [H ] remains unchanged since it is determined by the constant K w d. The answer cannot be determined without knowing [OH - ] e. [H ] starts fluctuating (i.e. sometimes it is lower at 55 o C than at 5 o C and sometimes higher) because of the increased thermal noise Answer: a. See Table 9-3 at the end of the exam booklet. 13. The dissociation of oxalic acid can be represented as: H OH CO HCO H O CO H O 3 3 Please write down the charge balance conditions for this equilibrium. a. b. c. d. e. [HCO ] [CO ] [OH ] = 3[H O ] 3 = 3 3 = 3 = 3 [HCO ] [CO ] [OH ] [H O ] [HCO ] [CO ] = [H O ] [HCO ] [CO ] [H O ] [HCO ] [CO ] 3[H O ] Answer: e. Just collect all negative charges on the left and all positive charges on the right. 1. The reaction leading to production of sulfur trioxide is described by the following equilibrium constant: [ SO3 ] K = [ X][ O ] Here X is the undisclosed molecule. Pick the reaction corresponding to this equilibrium constant: a. S O SO b. SO3 S 3O c. S 3 O SO3 O SO SO SO O SO d. 3 e. 3

7 Answer: d 15. One liter of buffer is prepared by mixing a solution of weak acid, HA, and its salt, NaA. At optimal conditions, ph pk a, the buffering capacity of the obtained solution is determined to be β=0.01. In the next step, we add a small amount of HCl to this buffer solution (specifically, 1ml of 1M solution of HCl). Estimate the change in ph caused by addition of HCl: a ph units b ph units c ph units d ph units e ph units dchcl chcl Answer: c. Use definition of β, β = dph ph 16. A 5.0-ml sample containing unknown concentration of Cu Cu ions, c 0, gave an absorbance readout A = 3.6 units (corrected for a blank). When exactly 0.5 ml of 0.087M Cu(NO 3 ) was added to the solution, the absorbance increased to 37.9 units. Cu Determine the original molar c 0 concentration assuming that the absorbance was directly proportional to the analyte concentration. a M b M c M d M e M Answer: d. Absorbance is proportional to concentration, A= βc. Cu N0 Before addition: 3.6 = β After addition: 37.9 N Cu = β Cu Where N 0 is the original amount of Cu Cu Cu, mol, and c0 = N0 / In a dilute solution of HCl the concentrations [H ] and [Cl - ] are related as [ H ]/[ Cl ] = What is the analytical concentration of HCl, c HCl, used to prepare this solution? The temperature is 5 o C.

8 a M b M c. 3.1 mm d M e M chcl chcl Kw Answer: e. Plug different answers into [ H ] = to determine [ H ], then recall that [Cl - ]= c HCl and verify that [ H ]/[ Cl ] = Two base molecules, pyruvate ion (conjugate base of pyruvic acid) and ammonia (conjugate base of ammonium ion) are competing for one and the same proton. Which one is more likely to win? a. pyruvate ion b. ammonia c. tie d. none of these molecules can bind a proton e. the outcome is random Answer: b. Once ammonia grabs the proton and forms NH, NH is much less likely to release the proton than pyruvic acid (cf. Ka values). Alternatively, calculate and compare Kb values. 19. If a molecule X is known to have a molar absorptivity of.0x10 3 M -1 cm -1 at 50 nm, and a solution of this compound showed a transmittance of 80% at this wavelength in a spectrophotometer with a path length of 1 cm, what is the concentration of X? Answer: b. a. 3.x10-3 M b..8x10-5 M c. 9.5x10 - M d. 7.9x10-5 M e. 3.5x10 - M P log = εbc P0 = log(0.8) (.0 10 )(1 )( ) 5 C x M =.8 10 x M cm cm C

9 0. On March 1 st, Alice made a sample consisting of barium iodate in water. After a week, the system equilibrated as shown in the plot A. In this graph, solid Ba(IO 3 ) at the bottom of the beaker is colored orange, while Ba ions in water are represented by an orange circle. On March 7 th, Bob came along. He measured the concentration of Ba ions in water in Alice s sample and found [Ba ] = 0.7mM. After that, Bob added 1L of water to the sample and left it sitting on a bench. On March 1 th, Bob returned and measured the concentration of Ba again (plot B). What did he find? [FYI: the graphs are for illustrative purpose only do not try to read the answer off the graphs!] a. [Ba ] = mm b. [Ba ] = 0.07 mm c. [Ba ] = 0. mm d. [Ba ] = 0.7 mm e. [Ba ] = 7.0 mm Answer: d. By March 1 th the system equilibrated again. The equilibrium does not depend on initial conditions, only on the properties of the system (i.e. the K eq ). Hence, the concentration was the same, 0.7 mm. 1. Alice synthesized a new type of monoprotic acid, which she called monoburbulic acid. To find out its K a, she prepared a 0.01 M aqueous solution of monoburbulic acid and measured the ph of this solution. The ph turned out to be After that Alice did some calculations and determined that: a. b. c. d. e. K a K a K a K a K a = M = M = M = M = M

10 ph Answer: e. You can determine [H ] = 10, then plug it into cubic equation (given at the end of the exam). In this cubic equation you know [H ], K w, and c HA. After you plug all these values in, you are left with a simple linear equation for K a, which is easy to solve. Reference Section

11 Approximate results for solutions of acids in water: 1. Ka Ka Kc a HA 13 [ H ] = valid when cha > 10 / K a. [ H ] = Kc K valid when c > 100K a HA w HA a [ H ] = Kc a HA valid when cha > max{100 Ka,10 / Ka}. c c K [ ] valid when 10 H = HA HA w 3 Ka M Rigorous (cubic) expression for solution of acid in water: = 3 [ H ] Ka[ H ] ( Kc a HA Kw)[ H ] KwKa 0

Volume NaOH Delivered (ml)

Volume NaOH Delivered (ml) Chemistry Spring 011 Exam 3: Chapters 8-10 Name 80 Points Complete five (5) of the following problems. Each problem is worth 16 points. CLEARLY mark the problems you do not want graded. You must show your