Lab 5 Report Thermodynamics

NAME: QUIZ SECTION: DATA AND CALCULATIONS Note: All sections of this report must be typed Part I. Flame emission determination of [K+] in the 0.50 M KCl + KHP (sat) solution Preparation and Analysis of Standards Table 1: Volume and AES Intensity Data for the Calibration Standards Standard [K +], (M) 1 2 3 4 5 7.50E-05 1.50E-04 2.25E-04 3.00E-04 3.75E-04 Volume of 5.00x10-3 M needed to make 50 mL of each calibration solution (mL) 0.75 1.50 2.25 3.00 3.75 Intensity 104 156 233 297 372 Intensity vs. [K+] 400 350 300 y = 902667x + 29.3 R² = 0.9967 Intensity 250 200 150 100 50 0 0.00E+00 5.00E-05 1.00E-04 1.50E-04 2.00E-04 2.50E-04 3.00E-04 3.50E-04 4.00E-04 [K+] (M) Calibration Summary: slope M-1 9.03E+05 y-intercept 29.3 Analysis of [K+] Saturated KHP in 0.50 M KCl Solution Table 2: Atomic Emission Data and Results for Determination of [K+] in the original sample Temperature, o C Intensity [K +] in Measured Solution (M) Dilution factor [K +] in Original Solution (M) 20.6 332 3.350E-04 2,500 0.838 Show your calculations for the [K+] in the measured and original solutions. measured = 332 = (9.03E5([K+]) ) + 29.3 [K+] = 332-29.3/9.03E5 = 3.35- 04M original solution: [K+]original = [K+]measured *2500 = (3.35E- 04)M(2500)=0.838 M Page 2 of 6 NAME: emily johnson QUIZ SECTION: AJ Part II. Determination of [HP-] in Saturated KHP Solutions A. Standardization of NaOH Table 3: Data and Results for Standardization of NaOH Run 1 0.132g Mass of KHP, g Molar Mass of KHP, g/mol Moles KHP 0.00065 Initial buret volume, mL 0.15 Final buret volume, mL 6.7 Volume NaOH titrated, mL 6.55 Volume NaOH titrated, L 0.00655 [NaOH], moles/L 0.099 Run 2 0.134g 204.22 Run 3** N/A 0.00066 0.25 6.75 6.5 0.0065 0.102 Average, [NaOH] 0.101 M Show your calculation for [NaOH] for Run #1. 0.00065 mol/0.00655 L= 0.0992 M -> 0.099 M Calculate the % difference between Runs #1 and #2. avg= (0.102M+0.099M)/2=0.101M (0.102-0.099)/0.101 = 0.0297 x 100 = 2.97% B. Titrations of KHP Samples With and Without KCl mL of KHP(aq) used in titrations of Solutions #1-6 1.00 mL Table 4: Titration Data for Solutions #1-6 Solution # Temperature (oC) Initial buret reading (mL) Final buret reading (mL) mL NaOH added 1 2 (trial 1) 3 4 5 6 0.4 20.7 33.9 42.7 52.2 57.4 0.30 0.20 4.55 0.40 0.35 0.30 2.8 4.55 9.60 6.25 6.05 7.60 2.50 4.35 5.05 5.85 5.70 7.30 Table 5: Data Processing for Solutions #1-6 Solution # Temperature (K) mmoles HP- [HP-] (M) Ksp 1 2 (trial 1) 3 4 5 6 273.6 293.9 307.1 315.9 325.4 330.6 0.253 0.439 0.510 0.591 0.576 0.737 0.253 0.439 0.510 0.591 0.576 0.737 0.0638 0.1930 0.2602 0.3491 0.3314 0.5436 Show your calculations for mmoles of NaOH, [HP-], and Ksp for Solution #2. mmoles NaOH: 0.101M x 4.35 mL = 0.439 mmoles [HP-]: mol/L = 0.439 mmol/1 mL = 0.439mol/1 L = 0.439 M Ksp: Ksp = [HP]^2 = (0.439M)^2 = 0.1927 -> 0.1930 Page 3 of 6 Conversions for creating the Van't Hoff plot: ln Ksp 1/T (K-1) 0.00366 0.00340 0.00326 0.00317 0.00307 0.00303 -2.753 -1.645 -1.346 -1.052 -1.104 -0.610 NAME: emily johnson QUIZ SECTION: AJ Evaluating the Temperature Dependence of Ksp P e A O Autofill of values for plot from previous page: ln(Ksp) vs. 1/T (K) 0.000 0.00290 0.00300 0.00310 0.00320 0.00330 0.00340 0.00350 0.00360 0.00370 -0.500 1/T (K-1) 0.00366 0.00340 0.00326 0.00317 0.00307 0.00303 ln Ksp -2.753 -1.645 -1.346 -1.052 -1.104 -0.610 ln(Ksp) -1.000 -1.500 y = -3059.9x + 8.5673 R² = 0.9474 -2.000 -2.500 -3.000 1/T (K) -3059.9 slope DHoSoln 25.4 kJ/mol y-intercept DSo Soln 8.567 R 8.314 J/mol*K 71.2 J/mol*K Show your calculations for DH and DS. DH = (-3059.9) (-8.314 J/mol *K)(10^-3 kJ/J)= 25.44 DS = (8.567)(8.314 J/mol*K)=71.2 J/mol*K Evaluating the Common Ion Effect on the Determination of Ksp Table 6: Replicate Titrations of Solution #2 (room temperature, no KCl) with NaOH(aq) to determine [HP-] Initial buret Final buret mL NaOH mmoles HP- mL KHP (aq) Solution # reading Temperature (oC) reading (mL) delivered used present (mL) 20.7 0.20 4.55 4.35 0.439 1.00 2 (trial 1) 1.00 2 20.7 0.35 4.90 4.55 0.460 1.00 2 21.6 2.85 7.03 4.18 0.422 Average: Table 7: Replicate Titrations of Solution #7 (room temperature, in KCl) with NaOH(aq) to determine [HP-] Autofill of [K+] from Table 2: o Solution # Temperature ( C) 7 7 7 21 21 21 Initial buret Final buret reading reading (mL) (mL) 0.15 3.08 4.90 7.55 0.05 2.85 mL NaOH delivered mmoles HPpresent mL KHP in KCl (aq) used 2.93 2.65 2.80 0.296 0.268 0.283 3.00 3.00 3.00 Average: [HP-] (M) Ksp 0.439 0.460 0.422 0.440 0.193 0.211 0.178 0.194 0.838 [HP-] (M) Ksp 0.099 0.089 0.094 0.094 0.083 0.075 0.079 0.079 Show your calculations for: 1) "Ksp in 0.50 M KCl" using the data in the first row of Table 7 and 2) the % difference between the two average Ksp values you obtained from Tables 6 and 7. Ksp: Ksp = [K+][HP-] = (0.838)(0.099) = 0.083 % difference: ((0.079-0.194)/((0.079+0.194)/2))x100=84.24% Page 4 of 6 NAME: QUIZ SECTION: RESULTS AND DISCUSSION 1. Why did you perform Atomic Emission analysis on the sample that contained both KHP and KCl? We performed atomic emission analysis on both KHP and KCl samples because we needed to determine the initial concentration of K+ ions in the solution before any KHP was added. 2. What is the purpose of the calibration curve? The calibration curve provides us with both a slope and y-intercept of the intensity vs concentration of K+, which allows us to determine the overall concentraton of K+ ions in both KHP and KCl solutions. With the calculated concentrations, we can then calculate Ksp. 3. Do we expect the two average Ksp values from Tables 6 & 7 above to be the same or different? Why? Discuss your results and comment the similarity/differences between the values you obtained. If the values are different than expected, explain possible reasons. Your average Ksp values from the previous page autofill here: Solution #2 (Table 6) 0.194 Solution #7 (Table 7) 0.079 We would expect the valyes from tables 6 and 7 to have some difference between them because in table 7 there is KCl and in table 6 there is no KCl. For table 6, the average Ksp value was 0.194 while for table 7 the averahe Ksp value was 0.079. These values are about 84% different. The difference can come from because of the lower concentrations od HP- due to more mL of added acid. Another possible reason for the difference may be due to experimental errors with the titrations. 4. Based on your results at room temperature (Solution #2), what is the solubility of KHP in water in g/L? Solution #2 (Table 6) Your average [HP-] from the previous page autofills here: 0.440 KHP solubility = ((0.440M HP-)(1.00E-3LKHP)(1 mol KHP/1 mol HP-)(204.22g/mol))/(1.00E-03 L KHP) = 89.856 g/L -> 89.9 g/L Page 5 of 6 M NAME: emily johnson QUIZ SECTION: AJ 5. What will happen to the solubility and Ksp when NaOH(aq) is added to a saturated solution of KHP? Explain your reasoning. When NaOH is added to the solubtion, the solubility will decrease. This is because tKHP is a weak acid reaction with a strong base, NaOH. The strong base neutralizes the hydronium ions within the solution, which would shift the equilibirum towards the forward durection. This will decrease the mols of KHP, and therefore the solubility would increase. The Ksp would remain constant because it is only dependent on temperature and not initial concentrations. 6. What will happen to the solubility and Ksp when HCl(aq) is added to a saturated solution of KHP? Explain your reasoning. The solubility will decrease due to the addition of HCl. HCl is a strong acid, which will increase the H+ ions in the solution. This increase of H+ ions would causeless KHP to dissolve in water and the solubility will be decreased. The Ksp would stay the same however, because it is only dependent on temperature and not concentrations. 7. Consider the following typical results: DHoSoln = 34.1 kJ/ mole and the DSoSoln = 120 J/K·mole. How do your results compare? Calculate % difference and explain possible reasons for differences. Your values from the ealier page autofill here: DH o 25.4 kJ/mol Soln DS o Soln 71.2 J/mol*K % difference dH(soln) = ((25.4 kJ/mol-34.1 kJ/mol)/(34.1kJ/mol+25.4kJ/mol)/2)x100 = 29.24% % difference dS(soln) = ((71.2 J/mol*K-120J/mol*K)/(120J/mol*K+21.2J/mol*K)/2)x100= 51.04% Our calculated dH(soln) value was 25.4 kJ/mol, which has a 29% difference than the actual value of 34.1 kJ/mol. Our calculated dS(soln) value was 71.2 J/mol*K, with a 51% difference from the typical value of 120 J/mol*K. This percent error can be attributed to mistakes in the titrations that gave inaccurate values. For example, there were times when the solution went from no color to bright pink in one drop. This could have given a wrongful recording of the NaOH used, which would then give a wrong calculation for the rest of the variables and values. Laboratory Waste Evaluation Laboratory waste is considered anythinggenerated during an experiment that is disposed of down the sewer drain, thrown in the garbage, collected in a container for disposal by the UW Environmental Health & Safety department, or released into the environment. Based on the written lab procedure andyour actions during the lab, list the identity and approximate amount (mass or volume) of waste that you generated while performing this experiment. 3 paper towels 6 7 pipette tips 2 weigh sheets 2 pH strips 75 mL 0.005M KCl and KHP 750 mL DI water 160 mL 0.1M NaOH 0.15 g KHP 2 mL phenolpthalein indicator Page 6 of 6