AP chem. Review – 16 Name __________________ (Experimental chemistry) 1. Qualitative analysis (2005, Question 5) Answer the following questions that relate to laboratory observations and procedures. a) An unknown gas is one of three possible gases: nitrogen, hydrogen, or oxygen. For each of the three possibilities,

describe the result expected when the gas is tested using a glowing splint (a wooden stick with one end that has been ignited and extinguished, but still contains hot, glowing, partially burned wood).

b) The following three mixtures have been prepared: CaO plus water, SiO2 plus water, and CO2 plus water. For each

mixture, predict whether the pH is less than 7, equal to 7, or greater than 7. Justify your answers.

c) Each of three beakers contains 0.1 M solution of one of the following solutes: potassium chloride, silver nitrate, or sodium sulfide. The three beakers are labeled randomly as solution 1, solution 2, and solution 3. Shown below is a partially completed table of observations made of the results of combining small amounts of different pairs of the solutions.

i. Write the chemical formula of the black precipitate.

ii. Describe the result of mixing solution 1 with 3.

iii. Identify each of the solutions 1, 2, and 3.

2. Qualitative analysis (2004, Question 5) In a laboratory class, a student is given three flasks that are labeled Q, R, and S. Each flask contains one of the following solutions: 1.0 M Pb(NO3)2, 1.0 M NaCl, or 1.0 M K2CO3. The student is also given two flasks that are labeled X and Y. One of these flasks contains 1.0 M AgNO3, and the other contains 1.0 M BaCl2. This information is summarized in the digram. a) When samples of Q and X are mixed, a precipitate formed. A precipitate also formed when samples of Q and Y are

combined. i. Identify solution Q.

ii. Write the chemical formulas for each of the two precipitates.

b) When solution Q is mixed with solution R, a precipitate forms. However, no precipitate forms when solution Q is mixed

with soluiton S. i. Identify R and S.

ii. Write the chemical formula of the precipitate that forms when Q is mixed with R.

c) X and Y are to be determined using only: 1.0 M solutions of Pb(NO3)2, NaCl, and K2CO3. i. Describe a procedure to identify solution X and solution Y.

ii. Describe observations that allow you to distinguish between X and Y.

iii. Explain how the observations enable you to distinguish between X and Y.

3. Beer’s law (2003, Question 5) A student is instructed to determine the concentration of a solution of CoCl2 based on absorption of light (spectrometric / colorimetric method). The student is provided with a 0.10 M solution of CoCl2 with which to prepare standard solutions with concentrations of 0.020 M, 0.040 M, 0.060 M, and 0.080 M. a) Describe the procedure for diluting the 0.10 M solution to a concentration of 0.020 M using distilled water, a 100 mL

volumetric flask, and a pipet or buret. Include specific amounts where appropriate.

The student takes the 0.10 M solution and determines the percent transmittance and the absorbance at various wavelengths. The two graphs represent the data.

b) Identify the optimum wavelength for the analysis. The student measures the absorbance of the 0.020 M, 0.040 M, 0.060 M, 0.080 M and 0.10 M solutions. The data are plotted below. c) The absorbance of the unknown solution is 0.275. What is the

concentration of the solution?

d) Beer’s Law is an expression that includes three factors that determine the amount of light that passes through a solution. Identify two of these factors.

e) The student handles the sample container (e.g. test tube or cuvette) that holds the unknown solution and leaves fingerprings in the path of the light beam. How will this affect the calculated concentration of the unkonwn? Explain your answer.

f) Why is this method of determining the concentration of CoCl2 solution appropriate whereas using the same method for measuring the concentration of NaCl solution would not be appropriate?

4. Beer’s law (2006, Form B, Question 5) A student carries out an experiment to determine the equilibrium constant for a reaction by colorimetric (spectrophotometric) analysis. The production of the red-colored species FeSCN2+(aq) is monitored. a) The optimum wavelength for the measurement of [FeSCN2+] must first be determined. The

plot of absorbance, A, versus wavelength, λ, for FeSCN2+(aq) is shown. What is the optimum wavelength for this experiment? Justify your answer.

b) A calibration plot for [FeSCN2+] is prepared at the optimum wavelength. The data give the

absorbances measured for a set of solutions of known concentration of FeSCN2+(aq). i. Draw a Beer’s law calibration plot of all the data on the grid. Add a scale on the horizontal axis.

ii. An FeSCN2+(aq) solution of unknown concentration has an absorbance of 0.300. Use the plot

you drew in part (i) to determine the concentration, in moles per liter, of this solution. c) The purpose of the experiment is to determine the equilibrium constant for the reaction:

Fe3+(aq) + SCN-(aq) ⇌ FeSCN2+(aq) i. Write the equilibrium-constant expression for Kc.

ii. The student combines solutions of Fe(NO3)3 and KSCN to produce a solution in which

the initial concentrations of Fe3+(aq) and SCN-(aq) are both 6.0 x 10-3 M. The absorbance of this solution is measured, and the equilibrium FeSCN2+(aq) concentration is found to be 1.0 x 10-3 M. Determine the value of Kc.

d) If the student’s equilibrium FeSCN2+(aq) solution of unknown concentration fades to a lighter color before the student

measures its absorbance, will the calculated value of Kc be too high, too low, or unaffected? Justify your answer.

5. Acid base titrations (2002, Form B, Question 2) The graph below shows the result of the titration of a 25 mL sample of a 0.10 M solution of a weak acid, HA, with a strong base, 0.10 M NaOH.

a) Describe two features of the graph above that identify HA as a weak acid. b) Describe one method by which the value of the acid-base dissociation constant for HA can be determined using the graph

above. c) On the graph above, sketch the titration curve that would result if 25 mL of 0.10 M HCl were used instead of 0.10 M HA. d) A 25 mL sample of 0.10 M HA is titrated with 0.20 M NaOH.

i. What volume of base must be added to reach the equivalence point?

ii. The pH at the equivalence point for this titration is slightly higher than the pH at the equivalence point in the titration using 0.10 M NaOH. Explain.

6. Electrochemical cells (2002, Form B, Question 7) The diagram below shows the experimental setup for a typical electrochemcial cell that contains two standard half-cells. The cell operates according to the reaction represented by the following equation.

a) Identify M and M2+ in the diagram and specify the initial concentration for M2+ in solution. b) Indicate which of the metal electrodes is the cathode. Write the balanced equation for the reaction that occurs in the half-

cell containing the cathode. c) What would be the effect on the cell voltage if the concentration of Zn2+ was reduced to 0.100 M in the half-cell containing

the Zn electrode? d) Describe what would happen to the cell voltage if the salt bridge was removed. Explain.

7. Kinetics (2005, Form B, Question 3) The decomposition of gas X to produce gases Y and Z is represented by: X → 2Y + Z. In a certain experiment, the reaction took place in a 5.00 L flask at 428 K. Use the information below to answer the questions which follow.

a) How many moles of X were initially in the flask? b) How many molecules of Y were produced in the first 20. minutes of the reaction? c) What is the order of this reaction with respect to X? Justify your answer. d) Write the rate law for this reaction. e) Calculate the specific rate constant for this reaction. Specify units. f) Calculate the concentration of X in the falsk after a total of 150. minutes of reaction. ANSWERS 1. 2005, Question 5 a) N2 will extinguish the splint, H2 will make a popping sound, O2 will make the splint glow brighter and reignite. b) CaO + H2O → Ca(OH)2, basic solution, pH > 7; SiO2 + H2O → insoluble, pH = 7; CO2 + H2O → H2CO3, acidic solution, pH < 7 c) (i) the black precipitate is Ag2S, (ii) a precipitate will be produced when the two solutions are mixed, (iii) 1 = AgNO3, 2 = Na2S, 3 = KCl 2. 2004, Question 5 a) (i) solution Q is K2CO3, (ii) Ag2CO3 and BaCO3 b) (i) solution R is Pb(NO3)2 and solution S is NaCl, (ii) PbCO3 c) (i) The identities of solutions X and Y can be determined by adding a sample of NaCl OR a sample of Pb(NO3)2 to each solution, (ii) Adding NaCl would form a white precipitate when added to AgNO3 but form no precipitate when added to BaCl2 OR adding Pb(NO3)2 would form a white precipitate when added to BaCl2 but will form no precipitate when added to AgNO3. (iii) When NaCl is added to X, a precipitate (AgCl) indicates that X is AgNO3. If no precipitate forms, then X is BaCl2. The same logic can be used to identify solution Y. OR if Pb(NO3)2 is added to solution X, a precipitate (PbCl2) indicates solution X is BaCl2. If no precipitate forms, X is AgNO3. The same logic can be used to identify solution Y. 3. 2003, Question 5 (see “Spectral Analysis” in appendix 3 of Zumdahl, 7th ed., p A16) a) Using M1V1 = M2V2 (the dilution equation), we know that (0.10)(V1) = (0.020)(100) and V1 = 20 mL. Pipet 20 mL of 0.10 M CoCl2 into the 100 mL volumetric flask, then add enough water to reach the 100 mL mark on the neck of the volumetric flask. Stopper the flask and mix. b) 510 nm (acceptable range 490-520 nm) c) 0.050 M (acceptable range 0.045 to 0.055 M) d) Beer’s law is absorbance = A = a·∙b·∙c where a = molar absorptivity, b = path length of cuvette/test tube, and c = concentration. e) Fingerprints will scatter or absorb light. Since less light reaches the detector, the solution will have a higher apparent absorbance, and therefore a higher reported concentration. f) A CoCl2(aq) solution absorbs visible light. A NaCl(aq) solution is colorless and does not absorb visible light. 4. 2006, Form B, Question 5 a) The optimum wavelength is 450 nm because that is the wavelength of maximum absorbance by FeSCN2+(aq). b) (i) create a linear plot and label the horizontal axis correctly, (ii) See plot in part (i). At A = 0.300, [FeSCN2+] = 16 x 10-4 mol/L. c) (i) Kc = [FeSCN2+] / [Fe3+][SCN-] (ii) Fe3+ + SCN- ⇌ FeSCN2+ I. 6.0 x 10-3 6.0 x 10-3 0 C. -1.0 x 10-3 -1.0 x 10-3 +1.0 x 10-3 E. 5.0 x 10-3 5.0 x 10-3 1.0 x 10-3 Kc = (1.0 x 10-3) / (5.0 x 10-3)2 = 40. d) The value of Kc will be too low; the lower absorbance reading indicates a lower [FeSCN2+] than actually existed before the fading occurred, so substitution of a lower [FeSCN2+] into the equilibrium expression will result in a lower value of Kc. 5. 2002, Form B, Question 2 a) (i) The initial pH is 3 (a strong acid would have a lower pH), (ii) The equivalence point is basic (above 7), (iii) There is a jump in pH before hitting the buffer range. b) At the maximum buffer capacity, at approximately 12.5 mL of NaOH added, the pH = pKa of the acid. Thus, pKa = 5, and Ka = 10-5. c) If 0.10 M HCl were used, the curve would start at pH = 1 and reach equivalence at pH = 7, 25 mL NaOH. d) (i) (0.10)(25) = (0.20)(X)… X = 12.5 mL NaOH, (ii) At equivalence, there are 2.5 moles of A-, the conjugate base of HA, regardless of using 25 mL 0.10 M NaOH or 12.5 mL of 0.20 M NaOH. However, since there is a smaller total volume when using 0.20 M NaOH, there is a greater concentration of A-, resulting in a more basic solution and a higher pH. When using 0.20 M: 12.5 mL + 25.0 mL = 37.5 mL and [A-] = (2.5/37.5) = 0.067 M. When using 0.10 M: 25.0 mL + 25.0 mL = 50.0 mL and [A-] = (2.5/50) = 0.050 M. 6. 2002, Form B, Question 7 a) M is Zn, and M2+ is Zn2+. The initial concentration of Zn2+ is 1.0 M because it says it operates under “standard conditions.” b) Nickel is the cathode. Ni2+(aq) + 2e- → Ni(s) c) If [Zn2+] is lowered to 0.100 M, the equilibrium for the net reaction would shift to the right, making the reaction more spontaneous and increasing the voltage. d) If the salt bridge was removed the voltage would drop to zero. The salt bridge allows ions to migrate and compensate for the buildup of Zn2+ ions on the left and the depletion of Ni2+ ions on the right. The cell cannot operate without a path for ion migration. 7. 2005, Form B, Question 3 a) [X] at 0 minutes = 0.00633, so 5.00 L x 0.00633 M = 3.17 x 10-2 mol X b) After 20. minutes of reaction, the number of moles of X remaining in the flask is (5.00 L)(0.00427 M) = 2.14 x 10-2 mol X. Then the number of moles of X that reacted in the first 20 minutes is (3.17 x 10-2 mol X) – (2.14 x 10-2 mol X) = 1.03 x 10-2 mol X. Thus the number of molecules of Y produced is (1.03 x 10-2 mol X)(2 mol Y/1 mol X)(6.02 x 1023 molecules Y/mol Y) = 1.24 x 1022 molecules Y. c) 1st order since ln[X] versus time is linear d) rate = k[X]1 e) ln [X]t/[X]0 = -kt… ln(0.00520/0.00633) = -k(10 min) and k = 0.0197 min-1. f) ln[X]150 – ln[X]0 = -kt… ln[X]150 = -(0.0197)(150) + ln(0.00633) = -8.017… e-8.017 = 3.30 x 10-4 M