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- Analytical Chem Test Notes.docx

Chelsea A.

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Log and Antilog n=10^a means logn=a log(5.4033E-8)=-7.2674 [num of sig figs in the log x = num after decimal] 10^6.124=1.39 [the num of sig figs behind the dec=num sig figs in whole ans] Error Systematic/determinate Random/indeterminate Absolute used in addition Sqrt(e^2+e^2) Relative percent used in mult Percent rel=Abs/mag of measure Ex percent rel unc. of buret 12.35 +/- 0.002 mL is =0.02/12.35 *100 Real Rule The first digit of the absolute uncertainty is the last significant digits in the answer. Ex 0.0946(+/-0.0002) St dev- the smaller the standard deviation of more closely the data are clustered about the mean. An experiment that produces a small standard deviation is more precise than one that produces a large standard deviation. Greater precision does not mean greater accuracy, which means closer to the truth. Students T: used to express the CL. 50 % is larger than 95%. Tcalc>Ttable the difference is significant Qcalc=gap/range S3-1. Indicate how many significant figures there are in: (a) 0.305 0 (b) 0.003 050 (c) 1.003 × 104 S3-2. Round each number as indicated: (a) 5.124 8 to 4 significant figures (d) 0.135 237 1 to 4 significant figures (b) 5.124 4 to 4 significant figures (e) 1.525 to 3 significant figures (c) 5.124 5 to 4 significant figures (f) 1.525 007 to 3 significant figures S3-3. Write each answer with the correct number of digits: (a) 3.021 + 8.99 = 12.011 (d) 0.030 2 ÷ (2.114 3 × 10-3) = 14.283 69 (b) 12.7 – 1.83 = 10.87 (e) log (2.2 × 10-18) = ? (g) 10-4.555 = ? (c) 6.345 × 2.2 = 13.959 0 (f) antilog (–2.224) = ? S3-4. Using the correct number of significant figures, find the formula mass of C6H13B. S3-5. Find the absolute and percent relative uncertainty and express each answer with a reasonable number of significant figures. (a) 3.4 (±0.2) + 2.6 (±0.1) = ? (c) [3.4 (±0.2) × 10-8] ÷ [2.6 (±0.1) × 103] = ? (b) 3.4 (±0.2) ÷ 2.6 (±0.1) = ? (d) [3.4 (±0.2) – 2.6 (±0.1)] × 3.4 (±0.2) = ? S3-6. Express the molecular mass (± uncertainty) of benzene, C6H6, with the correct number of significant figures. S3-7. (a) A solution is prepared by dissolving 0.222 2 (±0.000 2) g of KIO3 [FM 214.001 0 (±0.000 9)] in 50.00 (±0.05) mL. Find the molarity and its uncertainty with an appropriate number of significant figures. (b) Would the answer be affected significantly if the reagent were only 99.9% pure? S3-8. Find the absolute and percent relative uncertainty and express each answer with a reasonable number of significant figures. (a) 3.4 (±0.2) = ? (c) 103.4 (±0.2) = ? (e) log [3.4 (±0.2)] = ? (b) [3.4 (±0.2)]2 = ? (d) e3.4 (±0.2) = ? (f) ln [3.4 (±0.2)] = ? S3-9. The value of Boltzmann's constant (k) listed on the inside front cover of the book is calculated from the quotient R/N, where R is the gas constant and N is Avogadro's number. If the uncertainty in R is 0.000 070 J/(mol . K) and the uncertainty in N is 0.000 003 6 × 1023/mol, find the uncertainty in k. S3-10. Find the uncertainty in the molecular mass of B10H14 and write the molecular mass with the correct number of significant figures. CHAPTER 4: SUPPLEMENTARY PROBLEMS 5 STATISTICS S4-1. Consider Rayleigh's data for the mass of gas from air in Table 4-3. Find the (a) mean (b) standard deviation (c) variance S4-2. Suppose that a Gaussian population of measurements has a mean of 1 000 and a standard deviation of 50. What fraction of the population lies in the following intervals: (a) >1 000 (b) 950-1 050 (c) 850-1 150 (d) <900 (e) 930-1 030 (f) 912-991 S4-3. Write the equation of the smooth Gaussian curve in Figure 4-1. (Since the curve represents the results of 4 768 measurements, and each bar on the graph corresponds to a 20-h interval, you must use a factor of 4 768 × 20 in the numerator of Equation 4-3.) Use the equation to calculate the value of y when x = 1000 h and see if your calculated value agrees with the value on the graph. S4-4. Find the 95 and 99% confidence intervals for the mean mass of nitrogen from chemical sources in Table 4-3. S4-5. Two methods were used to measure the specific activity (units of enzyme activity per milligram of protein) of an enzyme. One unit of enzyme activity is defined as the amount of enzyme that catalyzes the formation of one micromole of product per minute under specified conditions. Enzyme activity (five replications) Method 1: 139 147 160 158 135 Method 2: 148 159 156 164 159 Is the mean value of method 1 significantly different from the mean value of method 2 at the 95% confidence level? S4-6. It is known from many careful measurements that the concentration of magnesium in a material is 0.137 wt %. Your new analytical procedure gives values of 0.129, 0.133, 0.136, 0.130, 0.128 and 0.131 wt %. Do your results differ from the expected result at the 95% confidence level? S4-7. Calcium in a mineral was analyzed five times by each of two methods, with similar standard deviations. Are the mean values significantly different at the 95% confidence level? Method 1: 0.027 1 0.028 2 0.027 9 0.027 1 0.027 5 Method 2: 0.027 1 0.026 8 0.026 3 0.027 4 0.026 9 sChapter 4: Supplementary Problems 6 S4-8. The Ti content (wt %) of two different ore samples was measured several times by the same method. Are the mean values significantly different at the 95% confidence level? Sample 1: 0.013 4 0.013 8 0.012 8 0.013 3 0.013 7 Sample 2: 0.013 5 0.014 2 0.013 7 0.014 1 0.014 3 S4-9. The calcium content of a person's urine was determined on two different days: Day Average Ca (mg/L) Number of measurements 1 238 4 2 255 5 The analysis applied to many samples yields a standard deviation of 14 mg/L. Are the two averages significantly different at the 95% confidence level? S4-10. Using the Q test, decide whether the value 0.195 should be rejected from the set of results: 0.217, 0.224, 0.195, 0.221, 0.221, 0.223. S4-11. (a) The table below lists rainfall measured in Los Angeles. Enter this data into a spreadsheet and compute the average and standard deviation. (b) Prepare a barchart showing rainfall as a function of year from 1878 to 1996. (c) Prepare a histogram (another barchart) from this data showing rainfall in 2-inch intervals in the format below. For example, the bar at x = 15 will show the number of years in which the rainfall was in the range 14.00 to 15.99 inches. In your judgement, does rainfall follow a Gaussian distribution? 0 S4-12. Students at Eastern Illinois University intended to prepare copper(II) carbonate by adding a solution of CuSO4 . 5H2O to a solution of Na2CO3. CuSO4.5H2O(aq) + Na2CO3(aq) → CuCO3(s) + Na2SO4(aq) + 5H2O(l) copper(II) carbonate After warming the mixture to 60°C, the gelatinous blue precipitate coagulated into an easily filterable pale green solid. The product was filtered, washed, and dried at 70° C. Copper in the product was measured by heating 0.4 g of solid in a stream of methane at high temperature to reduce the solid to pure Cu, which was weighed. 4CuCO3(s) + CH4(g) heat → 4Cu(s) + 5CO2(g) + 2H2O(g) In 1995, 43 students found a mean value of 55.6 wt % Cu with a standard deviation of 2.7 wt %. In 1996, 39 students found 55.9 wt % with a standard deviation of 3.8 wt %. The instructor tried the experiment 9 times and measured 55.8 wt % with a standard deviation of 0.5 wt %. Was the product of the reaction probably CuCO3? Could it have been a hydrate, CuCO3 . xH2O? [This problem was taken from D. Sheeran, J. Chem. Ed. 1998, 75, 453. See also H. Gamsjäger and W. Preis, J. Chem. Ed. 1999, 76, 1339.] CHAPTER 3: SUPPLEMENTARY SOLUTIONS 4 EXPERIMENTAL ERROR S3-1. (a) 4 (b) 4 (c) 4 S3-2. (a) 5.125 (b) 5.124 (c) 5.124 (d) 0.135 2 (e) 1.52 (f) 1.53 S3-3. (a) 12.01 (c) 14 (e) –17.66 (g) 2.79 × 10-5 (b) 10.9 (d) 14.3 (f) 5.97 × 10-3 S3-4. 95.978 S3-5. (a) 3.4 ± 0.2e = 0.22 + 0.12 = 0.224 + 2.6 ± 0.1 6.0 ± e = 6.0 ± 0.2 (±3.7%) (b) 3.4 ± 0.2 2.6 ± 0.1 = 3.4 ± 5.88% 2.6 ± 3.85% = 1.308 ±e %e = 5.882 + 3.852 = 7.03% Answer: 1.308 ± 0.092 (±7.0%) (c) 3 8 2.6( 0.1) 10 3.4( 0.2) 10 ± × ± × − = 1.30 ( 0.09 ) 10 ( 7. %) 2.6( 3.85%) 10 3.4( 5.88%) 10 0 11 3 8 2 8 = ± × ± ± × ± × − − (d) 3.4 (± 0.2) – 2.6 (±0.1) = 0.8 ± 0.224 = 0.8 ± 28.0% 0.8 (± 28.0%) × 3.4 (±5.88%) = 2.72 ± 28.6% Answer: 2.72 ± 0.78 (±29%) S3-6. C: 12.010 7 ± 0.000 8 H: 1.007 94 ± 0.000 07 +6C:6(12.010 7 ± 0.000 8) = 72.064 2 ± 0.004 8 +6H:6(1.007 94 ± 0.000 07) = 6.047 64 ± 0.000 42 ______________________________________________________________________________________ C6H6: 78.111 8 ± ? Uncertainty = 0.004 82 + 0.000 432 = 0.004 8 Answer: 78.112 ± 0.005 S3-7. (a) Molarity = 0.05000 ( 0.10%) L mol 214.0010 ( 0.000 42%) g 0.222 2 ( 0.090%) g ± / × ± ± / %e = 0.0902 + 0.000 422 + 0.102 = 0.135% molarity = 0.020 766 ± 0.000 028 M (b) The uncertainty in the analysis is ~0.1%, so 0.1% uncertainty in reagent purity is significant. Chapter 3: Supplementary Solutions 5 S3-8. (a) y = x1/2 ⇒ %ey = × 100 3.4 0.2 2 1 = 2.94% Answer: 1.844 ± 0.054 (±2.9%) (b) y = x2 ⇒ %ey = 2 × 100 3.4 0.2 = 11.76% Answer: 11.6 ± 1.4 (±12%) (c) y = 10x ⇒ ey = (103.4)(2.302 6) (0.2) = 1.16 × 103 Answer: 2.51 ± 1.16 × 103 (±46%) (d) y = ex ⇒ ey = (e3.4) (0.2) = 5.99 Answer: 30.0 ± 6.0 (±20%) (e) y = log x ⇒ ey = 0.434 29 0.2 3.4 = 0.025 5 Answer: 0.531 ± 0.026 (±4.8%) (f) y = ln x ⇒ ey = 0.2 3.4 = 0.058 8 Answer: 1.224 ± 0.059 (±4.8%) S3-9. k = RN ⇒ %e2k = %e2 R + %e2 N ⇒ %e2k = × 8.314 472 100 0.000070 2 + × 6.022 136 7 100 0.000 003 6 2 ⇒ %ek = 0.000 844% ⇒ ek = (0.000 008 44)(1.380 658) = 0.000 012 S3-10. B = 10.811 ± 0.007 H = 1.007 94 ± 0.000 07 +10B: 10(10.811 ± 0.007) = 108.110 ± 0.07 +14H: 14(1.007 94 ± 0.000 07) = 14.111 16 ± 0.000 98 ______________________________________________________________________________________ B10H143: 122.221 ± ? Uncertainty = 0.072 + 0.000 982 = 0.07 Answer: 122.22 ± 0.07 CHAPTER 4: SUPPLEMENTARY SOLUTIONS 6 S4-1. (a) Mean = 17 (2.310 17 + . . . + 2.310 28) = 2.310 11 (b) Standard deviation = σ = 2 1 / 2 6 ( ) Σ x − x i = 0.000 143 (c) Variance = σ2 = 2.03 × 10-8 S4-2. (a) z > 0 ⇒ 50% (c) z = –3 to z = +3 ⇒ 99.73% (b) z = –1 to z = +1 ⇒ 68.26% (d) z < –2 ⇒ 2.27% (e) z = –1.4 to z = +0.6 ⇒ area = 0.419 2 + 0.225 8 = 64.50% (f) z = –1.76 to z = –0.18 ⇒ area = 0.460 6 – 0.071 4 = 38.92% Interpolations: 1.76 – 1.70 1.80 – 1.70 (0.464 1 – 0.455 4) + 0.455 4 = 0.460 6 0.18 – 0.10 0.20 – 0.10 (0.079 3 – 0.039 8) + 0.039 8 = 0.071 4 S4-3. y = 94.2 2 p 4768 × 20 e–(x – 845.2)2/2(94.2)2 = 104.7 when x = 1 000. S4-4. –x = 2.299 47 g, s = 0.001 38 g, n = 7 degrees of freedom 95% confidence: μ = –x ± (2.365)(0.001 38) 8 = 2.299 47 ± 0.001 15 99% confidence: μ = –x ± (3.500)(0.001 38) 8 = 2.299 47 ± 0.001 71 S4-5. x1 = 147.8 , 2 2 x = 157. , spooled = 8.90, t = 5 5 5 5 8. 157. 147. 90 2 8 + − ⋅ = 1.67 < 2.306 (Student’s t for 8 degrees of freedom) The difference is not significant. S4-6. –x = 0.13117, s = 0.00293 tcalculated = n s | known value − x | = |0.137 – 0.13117| 0.00293 6 = 4.87 For 5 degrees of freedom and 95% confidence, ttable =2.571. Because tcalculated (4.87) > ttable (2.571), the difference is significant. STATISTICS Chapter 4: Supplementary Solutions 7 S4-7. For Method 1, we find 1 x = 0.027 56 and s1 = 0.000 488. For Method 2, 2 x = 0.026 90 and s2 = 0.000 406. Fcalculated = 0.000 4882/0.000 4062 = 1.44 < Ftable = 6.39 (for 4 degrees of freedom in both the numerator and denominator). Standard deviations are not significantly different at 95% confidence level. Because Fcalculated < Ftable, we can use Equations 4-8 and 4-9. spooled = s12 (n1 – 1) + s22 (n2 – 1) n1 + n2 – 2 = 0.000 4882 (5 – 1) + 0.000 4062 (5 – 1) 5 + 5 – 2 = 0.000 449 tcalculated = x - 1 – x - 2 spooled n1n2 n1 + n2 = 0.027 56 – 0.026 90 0.000 449 5 . 5 5 + 5 = 2.32 Because tcalculated (= 2.32) > ttable (= 2.306 for 8 degrees of freedom), the difference is significant at the 95% confidence level. S4-8. Sample 1: 1 00 x = 0.013 4 s1 = 0.000 3937 Sample 2: 2 60 x = 0.013 9 s2 = 0.000 3435 Fcalculated = 0.000 39372 / 0.000 34352 = 1.314 < Ftable = 6.39 (for 4 degrees of freedom in both the numerator and denominator). Standard deviations are not significantly different at 95% confidence level. Because Fcalculated < Ftable, we can use Equations 4-8 and 4-9. spooled = 4s21 + 4s22 5 + 5 – 2 = 0.000 3695 tcalculated = 0.013 960 – 0.013 400 0.000 3695 5 . 5 5 + 5 = 2.40 > 2.306 (Student's t for 8 degrees of freedom) The difference is significant. S4-9. t = 255 – 238 14 4.5 4+5 = 1.81 < 2.365 (Student's t for 7 degrees of freedom) Difference is not significant. Chapter 4: Supplementary Solutions 8 S4-10. Q = (0.217 – 0.195) / (0.224 – 0.195) = 0.76 > 0.56. Discard 0.195. S4-11. (a) average = 15.01 standard deviation = 6.89 (b) 0 10 20 30 40 Rainfall 1878 1881 1884 1887 1890 1893 1896 1899 1902 1905 1908 1911 1914 1917 1920 1923 1926 1929 1932 1935 1938 1941 1944 1947 1950 1953 1956 1959 1962 1965 1968 1971 1974 1977 1980 1983 1986 1989 1992 1995 Year (c) 0 5 10 15 Number of years 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 Inches of rain The distribution does not look Gaussian at all. Chapter 4: Supplementary Solutions 9 S4-12. The formula mass of CuCO3 is 123.555 and Cu in this formula is 51.43 wt %. For the 1995 class data, the 95% confidence interval is μ (95%) = –x ± ts n = 55.6 ± (2.02)(2.7) 43 = 55.6 ± 0.8 = 54.8 to 56.4 wt % The 99% confidence interval is μ (99%) = –x ± ts n = 55.6 ± (2.70)(2.7) 43 = 55.6 ± 1.1 = 54.5 to 56.7 wt % Even the 99% confidence interval does not include the Cu content in CuCO3 (51.43 wt %). From the 1996 class data, the 99% confidence interval is 54.3 to 57.5 wt %. From the instructor's measurements, the 99% confidence interval is 55.2 to 56.4 wt %. The product cannot be CuCO3. It cannot be a hydrate either, because CuCO3 . xH2O, would have an even lower Cu content than 51.43%. The observed composition is closer to that of the minerals azurite, Cu3(OH)2(CO3)2 (55.31 wt % Cu), or malachite, Cu2(OH)2(CO3) (57.48 wt % Cu), than it is to CuCO3 CHAPTER 6: SUPPLEMENTARY PROBLEMS 11 CHEMICAL EQUILIBRIUM S6-1. Write the expression for the equilibrium constant for each of the following reactions. Write the pressure of a gaseous molecule, X, as PX. (a) Cl2(g) + 2OH-(aq) Cl-(aq) + OCl-(aq) + H2O(l) (b) Hg(l) + I2(g) HgI2(s) S6-2. Suppose that the following reaction has come to equilibrium: Br2(l) + I2(s) + 4Cl-(aq) 2Br-(aq) + 2ICl2 - (aq) If more I2(s) is added, will the concentration of ICl2 - in the aqueous phase increase, decrease, or remain unchanged? S6-3. From the equations CuN3(s) Cu+ + N3 - K = 4.9 × 10-9 HN3 H+ + N3 - K = 2.2 × 10-5 find the value of K for the reaction Cu+ + HN3 CuN3(s) + H+. All species are aqueous unless otherwise indicated. S6-4. For the reaction H2O(l) H+(aq) + OH-(aq), K = 1.0 × 10-14 at 25°C. The concentrations in a system out of equilibrium are [H+] = 3.0 × 10-5 M and [OH-] = 2.0 × 10-7 M. Will the reaction proceed to the left or to the right to reach equilibrium? S6-5. For the sum of two reactions, we know that K3 = K1K2. Show that this implies that Δ ° = Δ ° + Δ °3 1 2 G G G A + B C + D K1 D + E B + F K2 A + E C + F K3 S6-6. Find ΔG° for the reactions (a) Ca(OH)2(s) Ca2+ + 2OH- K = 6.5 × 10-5 (b) Mg(OH)2(s) Mg2+ + 2OH- K = 7.1 × 10-12 S6-7. For the reaction Mg2+ + Cu(s) Mg(s) + Cu2+, K = 10-92 and ΔS° = +18 J/(K.mol). (a) Under standard conditions, is ΔG° positive or negative? The term "standard conditions" means that reactants and products are in their standard states. (b) Under standard conditions, is the reaction endothermic or exothermic? S6-8. Use the solubility product to calculate the solubility of Ag2CrO4 (FM 331.73) (→ 2Ag+ + CrO 4 2- ) in water expressed as (a) moles per liter, (b) g/100 mL and (c) ppm Ag+(≈ μg Ag+/mL). Chapter 6: Supplementary Problems 12 S6-9. The solubility product for CuCl is 1.9 × 10-7. The equilibrium constant for the reaction Cu(s) + Cu2+ 2Cu+ is 9.6 × 10-7. Calculate the equilibrium constant for the reaction Cu(s) + Cu2+ + 2Cl- 2CuCl(s) S6-10. How many grams of PbI2 (FM 461.0) will dissolve in 0.500 L of (a) water and (b) 0.063 4 M NaI? S6-11. What concentration of Ca2+ must be added to 0.010 M oxalate (C2O4 2- ) to precipitate 99.0% of the oxalate? S6-12. Is it possible to separate 99.90% of 0.020 M Mg2+ from 0.10 M Ca2+ without precipitation of Ca(OH)2 by addition of NaOH? S6-13. Using Equations 6-11 to 6-15, calculate the concentrations of Pb2+, PbI+, PbI2(aq), PbI3 - and PbI4 - in a solution whose total I- concentration is somehow fixed at 0.050 M. Compare your answers to Figure 6-2. S6-14. Consider the following equilibria: AgCl(s) Ag+ + Cl- Ksp = 1.8 × 10-10 AgCl(s) + Cl- AgCl2 - K2 = 1.5 × 10-2 AgCl2 - + Cl- AgCl 3 2- K3 = 0.49 Find the total concentration of silver-containing species in a silver-saturated, aqueous solution containing the following concentrations of Cl-: (a) 0.010 M (b) 0.20 M (c) 2.0 M S6-15. Identify the Brønsted-Lowry acids on both sides of the reaction NaHSO3 + NaOH Na2SO3 + H2O S6-16. Identify the conjugate acid-base pairs in the reaction H2 NCH2CH2NH2 + H2O H3N + CH2CH2NH2 + OHEthylenediamine S6-17. Calculate the concentration of H+ and the pH of: (a) 0.001 0 M HClO4 (d) 3.0 M NaOH (b) 0.050 M HBr (e) 0.005 0 M [(CH3CH2)4N+]OH- (c) 0.050 M LiOH tetraethylammonium hydroxide S6-18. Write the Ka reaction for formic acid, HCO2H, and for methylammonium ion, CH3NH+ 3 . Chapter 6: Supplementary Problems 13 S6-19. Write the Kb reactions for piperidine and benzoate. NH CO2 Piperidine Benzoate - S6-20. Write the Ka and Kb reactions of K2HPO4. S6-21. Write the stepwise acid-base reactions for the following species in water. Write the correct symbol (e.g., Kb1) for the equilibrium constant for each reaction. S6-22. Use Appendix G to decide which is the stronger acid: 3-nitrophenol or 4-nitrophenol. Write the acid dissociation reaction of each. S6-23. Which is the stronger base: cyclohexylamine or imidazole? Write the base hydrolysis reaction of each. In the case of imidazole, the nitrogen atom without a hydrogen is the one that accepts H+. NH2 N N H Cyclohexylamine Imidazole pKB= 3.36 pKB= 7.01 S6-24. Write the Kb reaction of hypochlorite, OCl-. Given that the Ka value for HOCl is 3.0 × 10-8, calculate Kb for OCl-. S6-25. Write the Ka2 reaction of H2SO4 and the Kb2 reaction the trisodium salt below. ONa ONa NaO S6-26. From the Ka values for citric acid in Appendix G, find Kb1, Kb2 and Kb3 for trisodium citrate S6-1. (a) K = [Cl-] [OCl-] / [OH-]2 PCl2 (b) K = 1/PI2 S6-2. Remain unchanged. S6-3. Cu+ + N-3 CuN3(s) K1 = 4.9 10 9 1 × − HN3 H+ + N-3 K2 = 2.2 × 10-5 __________________________________________________ Cu+ + HN3 CuN3(s) + H+ K3 = K1K2 = 4.5 × 103 S6-4. Q = [H+][OH-] = 6.0 × 10-12 > K ⇒ reaction goes to the left. S6-5. K3 = e -ΔG3 °/RT = K1 . K2 = e -ΔG1 °/RT e -ΔG2 ° /RT e -ΔG3 °/RT = e -(ΔG1 ° + ΔG2 °)/RT ∴ ΔG3 ° = ΔG1 ° + ΔG2 ° S6-6. K = e -ΔG°/RT ⇒ ln K = –ΔG°/RT ⇒ ΔG° = –RT ln K (a) ΔG° = – (8.314 5 J mol.K ) (298.15 K) ln (6.5 × 10-5) = 23.9 kJ/mol (b) ΔG° = 63.6 kJ/mol S6-7. (a) Positive. (b) ΔG° = (+) = ΔH° – T ΔS°. Since –T ΔS° is negative, ΔH° must be positive. The reaction is endothermic. S6-8. (a) Ag2CrO4(s) 2Ag+ + CrO2- 4 FM 331.73 2x x (2x)2(x) = 1.2 × 10-12 ⇒ x = 6.69 × 10-5 M (b) 6.69 × 10-5 M = 0.022 2 g/L = 0.002 22 g/100 mL (c) [Ag+] = 13.4 × 10-5 M = 0.014 4 g/L = 0.014 4 mg/mL = 14.4 μg/mL = 14.4 ppm S6-9. Cu(s) + Cu2+ 2 Cu+ K1 = 9.6 × 10-7 2 Cu+ + 2 Cl- 2 CuCl(s) K2 = 1/Ksp2 = 1/(1.9 × 10-7)2 ________________________________ Cu(s) + Cu2+ + 2 Cl- 2 CuCl(s) K = K1K2 = 2.7 × 107 Ksp Chapter 6: Supplementary Solutions 13 S6-10. (a) PbI2(s) Pb2+ + 2IFM 461.0 x 2x x (2x)2 = 7.9 × 10-9 ⇒ x = 1.25 × 10-3 M = 0.578 g/L (b) [Pb2+] (0.063 4)2 = 7.9 × 10-9 ⇒ x = 1.97 × 10-6 M = 9.06 × 10-4 g/L S6-11. Ca(C2O4)(s) Ca2+ + C2O2- 4 We want to reduce [C2O2- 4 ] to 1.0 × 10-4 M : [Ca2+] [1.0 × 10-4] = 1.3 × 10-8 ⇒ [Ca2+] = 1.3 × 10-4 M S6-12. Ca(OH)2: Ksp = 6.5 × 10-6 Mg(OH)2: Ksp = 7.1 × 10-12 We want to reduce [Mg2+] to 2.0 × 10-5 M [Mg2+] [OH-]2 = [2.0 × 10-5] [OH-]2 = 7.1 × 10-12 ⇒ [OH-] = 5.96 × 10-4 M Will this precipitate 0.10 M Ca2+? Q = [Ca2+] [OH-]2 = (0.10)(5.96 × 10-4)2 = 3.55 × 10-8 < Ksp ⇒ Ca(OH)2 will not precipitate. S6-13. [Pb2+] = Ksp/[I-]2 = (7.9 × 10-9) / (0.050)2 = 3.16 × 10-6 M [PbI+]= K1[Pb2+] [I-] = 1.58 × 10-5 M [PbI2(aq)] = β2[Pb2+] [I-]2 = 1.11 × 10-5 M [PbI-3 ] = β3[Pb2+] [I-]3 = 3.28 × 10-6 M [PbI2- 4 ] = β4[Pb2+] [I-]4 = 5.92 × 10-7 M S6-14. [Ag+] = Ksp/[Cl-] [AgCl2- 3 ] = K3[AgCl-2 ] [Cl-] [AgCl-2 ] = K2[Cl-] [Ag]total = [Ag+] + [AgCl-2 ] + [AgCl2- 3 ] (a) 0.010 M Cl- (b) 0.20 M Cl- (c) 2.0 M Cl- [Ag+] 1.80 × 10-8 M 9.00 × 10-10 M 9.00 × 10-11 M [AgCl-2 ] 1.50 × 10-4 M 0.003 00 M 0.0300 M [AgCl2- 3 ] 7.35 × 10-7 M 0.000 294 M 0.0294 M [Ag]total 1.50 × 10-4 M 0.003 29 M 0.0594 M S6-15. HSO-3 , H2O Ksp Ksp = 1.3 × 10-8 Chapter 6: Supplementary Solutions 14 S6-16. acid base H2O OHH3 +N CH2CH2NH2 H2NCH2CH2NH2 S6-17. (a) [H+] = 1.0 × 10-3 M ⇒ pH = –log [H+] = 3.00 (b) [H+] = 0.050 M ⇒ pH = 1.30 (c) [OH-] = 0.050 M ⇒ [H+] = Kw / [OH-] = 2.0 × 10-13 M ⇒ pH = 12.70 (d) [OH-] = 3.0 M ⇒ [H+] = 3.3 × 10-15 M ⇒ pH = 14.48 (e) [OH-] = 0.005 0 M ⇒ [H+] = 2.0 × 10-12 M ⇒ pH = 11.70 S6-18. HCO2H HCO-2 + H+ CH3NH+ 3 CH3NH2 + H+ S6-19. NH + H 2 O NH2 + OHK b + C O + H 2 O K b CO2H - + OH- 2 S6-20. − 24 HPO H+ + PO3- 4 HPO2- 4 + H2O H2PO-4 + O HS6- 21. K HN HN + - N H + H 2 O NH + OH b 1 N H2 + H2O K b 2 HN + H+ 2N NH2+ + OHC O 2 + H 2 O K b 2 - CO2H CO2 + OHCO 2 H C O2 + H2 O K b 1 - CO2H CO 2 - + OH - CO 2 - K Ka Ka Kb Chapter 6: Supplementary Solutions 15 S6-22. 4-nitrophenol has the larger Ka. NO 2 OH K a O- + H+ NO 2 O 2 N OH O O- + H+ 2 N K a S6-23. Cyclohexylamine has the larger Kb. NH 2 + H 2 O K b NH+3 + OHN H N + H 2 O K b + OHN+ H N H S6-24. OCl- + H2O HOCl + OH- Kb = Kw/Ka = 3.3 × 10-7 S6-25. HSO H+ + SO2- 4 OOH K b 2 + OH- O O- HO OH + H 2 O S6-26. Kb1 = Kw Ka3 = 2.49 × 10-8 Kb2 = Kw Ka2 = 5.78 × 10-10 Kb3 = Kw Ka1 = 1.34 × 10-11 Ka2 S9-1. Write a charge balance for an aqueous solution of glycine, which reacts as follows: +H3NCH2CO2 - H2NCH2CO2 - + H+ +H3NCH2CO2 - + H2O +H3NCH2CO2H + OHglycine S9-2. Write a charge balance for a solution of Al(OH)3 dissolved in 1 M KOH. Possible species are Al3+, AlOH2+, Al(OH)2 + , Al(OH)3 and Al(OH)4 - . S9-3. Write a mass balance for a 0.05 M solution of glycine (Problem S9-1) in water. S9-4. Suppose that 0.30 g of AlOOH (FM 59.99) plus 150 mL of 3.0 M KOH are diluted to l.00 L to give the same species produced by Al(OH)3 in Problem S9-2. Write mass balance equations for aluminum and potassium. S9-5. Use the systematic treatment of equilibrium to calculate the concentration of Hg 2 2+ in a saturated aqueous solution of (Hg2)3[Co(CN)6]2 which dissociates into mercurous ion and Co(CN) 6 3- (cobalticyanide) . S9-6. A solution is prepared by mixing Mt moles of the salt MCl2 (which dissociates completely to M2+ + 2Cl-) and Lt moles of the ligand HL in 1 L. The following reactions may occur: M2+ + L-) ML+ K = 1.0 × 108 HL(aq) L- + H+ Ka = 1.0 × 10-5 (a) Write a mass balance for the metal species. (b) Write a mass balance for the ligand species. (c) Write a charge balance. (d) Suppose that Mt = Lt = 0.1 M (exactly) and the pH is somehow fixed at 5.00. (This means that the charge balance no longer applies.) Use the equilibria and the two mass balances to find the concentrations of ML+, M2+, L-, and HL. S9-7. (a) Use the procedure in Section 9-4 to find the concentrations of Mg2+, F-, and HF in a saturated aqueous solution of MgF2 held at pH 3.00. (b) Look up the formation constant for MgF+ in the Appendix. Using the concentrations of Mg2+ and F- from (a), calculate the concentration of MgF+. Is it negligible compared to [Mg2+]? (The answer is no.) (c) Because [MgF+] is not negligible compared to [Mg2+], we need to alter the mass balance to solve this problem correctly. Write the mass balance including the species MgF+. S9-8. The acid HA has a solubility of 0.008 5 M in water at 25°C. Ks HA(s) HA(aq) Ks = [HA(aq)] = 0.008 5 (a) MONOPROTIC ACID-BASE EQUILIBRIA Chapter 9: Supplementary Problems 18 If NaOH is added to a suspension of solid HA in water, more acid dissolves because of the reaction K = 6.3 × 105 HA(aq) + OH- A-(aq) + H2O (b) Consider a saturated solution, whose pH is somehow fixed at 10.00, in contact with excess solid HA. Calculate the total concentration of HA + A-. S9-9. Consider a saturated solution of SrSO4 in which the following reactions can occur: SrSO4(s) Sr2+ + SO 4 2- Ksp = 3.2 × 10-7 SO4 2- + H2O HSO4 - + OH- Kb = 9.8 × 10-13 (a) Write mass and charge balances for this solution. (b) Find the concentration of Sr2+ in the solution if the pH is fixed at 2.50. S9-10. The previous problem neglected ion pair formation: Sr2+ + SO SrSO4(aq) K = 1.6 × 102 (a) Write the mass balance including the ion pair. (b) Find the concentrations of Sr2+ and SrSO4(aq) if the pH is fixed at 2.50. What fraction of dissolved strontium is in the ion pair? S9-11. Consider a saturated solution of calcium oxalate, CaC2O4, in which the following reactions can occur: CaC2O4(s) 4 2- Ca2+ + C2O4 2- Ksp = 1.3 × 10-8 C2 O4 2- + H2O HC2O4 - + OH- Kb1 = 1.8 × 10-10 HC2O4 - + H2O H2C2O4 + OH- Kb2 = 1.8 × 10-13 (a) Write mass and charge balances for this solution. (b) Find the concentration of Ca2+ in the solution if the pH is fixed at 2.30. S9-12. Consider a saturated solution of zinc arsenate, Zn3(AsO4)2, in which the following reactions can occur: Zn3(AsO4)2(s) 3Zn2+ + 2AsO 4 3- Ksp = 1.0 × 10-27 AsO 4 3- + H2O HAsO 4 2- + OH- Kb1 = 3.1 × 10-3 HAsO 4 2- + H2O H2AsO4 - + OH- Kb2 = 9.1 × 10-8 H2AsO4 - + H2O H3AsO4 + OH- Kb3 = 1.7 × 10-12 (a) Write mass and charge balances for this solution. (b) Find the concentration of Zn2+ in the solution if the pH is fixed at 6.00. S9-13. Use the data in Problem S9-6 to construct a graph showing the concentrations of M2+, ML+, L- and HL as a function of pH from 0 to 14 in 0.5 increments. CHAPTER 11: SUPPLEMENTARY SOLUTIONS 34 S11-1. CH 2 OH CH2OH H 2 N CH C O 2 + H 2 O H 3 N CH CO2 + OH - + - - Kb1 = Kw/Ka2 = 1.62 × 10-5 CH 2 OH CH2OH - + - H 3 N C H CO 2 + H 2 O H 3 N C HCO2H + OH + Kb2 = Kw/Ka1 = 1.54 × 10-12 S11-2. (a) x2 0.100 – x = K1 ⇒ x = 9.95 × 10-4 M = [H+] = [HA-] ⇒ pH = 3.00 [H2A] = 0.100 – x = 0.099 0 M [A2-] = K2 [HA-] [H+] = 1.00 × 10-9 M (b) [H+] = K1K2F + K1Kw K1 + F = 1.00 × 10-7 M ⇒ pH = 7.00 [HA-] ≈ 0.100 M [H2A] = [H+][HA-] K1 = 1.00 × 10-3 M [A2-] = K2[HA-] [H+] = 1.00 × 10-3 M (c) x2 0.100 – x = Kw K2 ⇒ x = [OH-] = [HA-] = 9.95 × 10-4 M ⇒ pH = 11.00 [A2-] = 0.100 – x = 0.099 0 M [H2A] = [H+][HA-] K1 = 1.00 × 10-9 M pH [H2A] [HA-] [A2-] 0.100 M H2A 3.00 9.90 × 10-2 9.95 × 10-4 1.00 × 10-9 0.100 M NaHA 7.00 1.00 × 10-3 0.100 1.00 × 10-3 0.100 M Na2A 11.00 1.00 × 10-9 9.95 × 10-4 9.90 × 10-2 S11-3. N NH+ Cl- (BH+ Cl-) (BH+ Cl-) is the intermediate form of a diprotic system, with F = 0.150 M , K1 = 4.65 × 10-6 and K2 = 1.86 × 10-10. [H+] = K1K2F + K1Kw K1 + F = 2.94 × 10-8 M ⇒ pH = 7.53 [BH+] ≈ 0.150 M [B] = K2[BH+] [H+] = 9.49 × 10-4 M ACID-BASE TITRATIONS Chapter 11: Supplementary Solutions 35 [BH 2 2+ ] = [BH+][H+] K1 = 9.48 × 10-4 M S11-4. Charge balance: [H+] + [Na+] + [BH+] = [Cl-] + [HA-] + 2[A2-] +[OH-] Mass balances: [Cl-] = F1 [Na+] = 2F2 F2 = [H2A] + [HA-] + [A2-] F3 = [B] + [BH+] Equilibria : [H+]γH+ [OH-] γOH- = Kw K1 = [H+]γH+ [HA-] γHA- / [H2A] γH2A K2 = [H+] γH+ [A2-] γA2- / [HA-] γHAKb = [BH+] γBH+[OH-] γOH- / [B]γB S11-5. Tartaric acid: H2A pK1 = 3.036 pK2 = 4.366 (a) At pH 3.00, there is a mixture of H2A and HA-. H2A + OH- → HAInitial mmol: 3.331 6 x — Final mmol: 3.331 6 – x — x 3.00 = 3.036 + log x 3.331 6 – x ⇒ x = 1.597 mmol = 3.77 mL KOH. (b) At pH 4.00, there is a mixture of HA- and A2-. We must add 3.331 6 mmol (= 7.876 mL) of KOH to convert H2A into HA-. Then we need to add more KOH: HA- + OH- → A2- + H2O Initial mmol: 3.331 6 x — Final mmol: 3.331 6 – x — x 4.00 = 4.366 + log x 3.331 6 – x ⇒ x = 1.003 mmol = 2.370 mL KOH Total KOH = 7.876 + 2.370 = 10.25 mL Chapter 11: Supplementary Solutions 36 S11-6. Malonic acid = H2A pK1 = 2.847 pK2 = 5.696 (a) At pH 6.00, there is a mixture of A2- and HA-. A2- + H+ → HAInitial mmol: 2.775 x — Final mmol: 2.775 – x — x 6.00 = 5.696 + log 2.775 – x x ⇒ x = 0.920 8 mmol = 2.19 mL HCl. (b) At pH 3.20, there is a mixture of H2A and HA-. We first add 2.775 mmol (= 6.591 mL) of HCl to convert A2- into HA-. Then we need to add more HCl HA- + H+ → H2A Initial mmol: 2.775 x — Final mmol: 2.775 – x — x 3.20 = 2.847 + log 2.775 – x x ⇒ x = 0.852 7 mmol = 2.025 mL HCl Total HCl = 6.591 + 2.025 = 8.62 mL S11-7. Oxalic acid = H2A pK1 = 1.252 pK2 = 4.266 At pH 3.20, there is a mixture of A2- and HA-. We begin with 5.00 g = 0.030 08 mol A2-. The reaction of H2A with A2- creates 2 moles of HA-: H2 A + A2- → 2HAInitial mmol: x 0.030 08 — Final mmol: — 0.030 08 – x 2x 3.20 = 4.266 + log 0.030 08 – x 2x ⇒ x = 0.025 67 mol = 2.31 g oxalic acid. S11-8. NH3 CH-CH2CO2 CO2 NH 2 CH-CH 2 CO 2 CO 2 NH 3 CH-CH 2 CO 2 H CO 2 H NH 3 CH-CH2CO2H CO 2 + + - - - - + K 1 K2 K3 - Aspartic acid Chapter 11: Supplementary Solutions 37 NH 3 CH-CH 2 CH 2 CH 2 NHC CO 2 H NH3 CH-CH2CH2CH 2 NHC CO2 NH 2 CH-CH 2 CH 2 CH 2 NHC CO 2 NH2 CH-CH2CH2CH 2 NHC CO2 K 3 K1 K 2 + N H 2 N H 2 + + N H 2 N H 2 + - - N H 2 N H 2 + N H 2 N H - Arginine S11-9. (a) Histidine pK1 = 1.7 pK2 = 6.02 pK3 = 9.08 H His+ HHis His - H 3 His 2 2+ For 0.050 0 M histidine, [H+] = K2K3F + K2Kw K2 + F = 2.82 × 10-8 M ⇒ pH = 7.55 7.55 = 6.02 + log [HHis] [H2His+] ⇒ [H2His+] [HHis] = 0.029 5 (b) For 0.050 0 M H2His+Cl-, [H+] = K1K2F + K1Kw K1 + F =1.17 × 10-4 M ⇒ pH = 3.93 3.93 = 6.02 + log [HHis] [H2His+] ⇒ [H2His+] [HHis] = 123 S11-10. NH3 CH-CH2CH2CH2NHC CO2 + NH2 NH2 + - H2Arg+ found in Arginine . HCl For H3Arg2+ pK1 = 1.823 pK2 = 8.991 pK3 = 12.48 [H+] = K1K2(0.012 0) + K1Kw K1 + (0.012 0) = 2.61 × 10-6 M ⇒ pH = 5.58 [H2Arg+][H+] [H3Arg2+] = K1 ⇒ [H3Arg2+] = 1.823 6 10 (0.012 0)(2.61 10 ) − × − = 2.08 × 10-6 M [HArg][H+] [H2Arg+] = K2 ⇒ [HArg] = 6 8.991 2.61 10 (10 )(0.012 0) × − − = 4.69 × 10-6 M Chapter 11: Supplementary Solutions 38 [Arg-][H+] [HArg] = K3 ⇒ [Arg-] = 6 12.48 6 2.61 10 (10 )(4.69 10 ) × − − × − = 5.95 × 10-13 M S11-11. (a) Since pK = 11.305, BH+ is predominant at pH 11 and B is predominant at pH 12. (b) 11.305 (c) 12.00 = 11.305 + log [B] [BH+] ⇒ [B] [BH+] = 5.0 2.00 = 11.305 + log [B] [BH+] ⇒ [B] [BH+] = 5.0 × 10-10 S11-12. (a) HSO3 - (b) HSO3 - (c) HSO3 - (d) SO 3 2- S11-13. H3Cit H2Cit- HCit2- Cit3- At pH 5.00, HCit2- is dominant. S11-14. Fraction in form HA = αHA = [H+] [H+] + Ka = 10-8.00 10-8.00 + 10-7.00 = 0.090 9 Fraction in form A- = αA- = Ka [H+] + Ka = 0.909 1. [A-] [HA] = 0.909 1 0.090 9 = 10.0. S11-15. Fraction in form BH 2 2+ = αBH 2 2+ = [H+]2 [H+]2 + [H+]K1 + K1K2 , where K1 = 10-6.00 and K2 = 10-12.00 ⇒ αBH 2 2+ = 9.98 × 10-4 S11-16. K1 = 3.8 × 10-5 K2 = 3.8 × 10-6 pH 5.00 pH 6.00 αH2A = [H+]2 [H+]2 + [H+]K1 + K1K2 = 0.16 0.005 4 αHA- = [H+]K1 [H+]2 + [H+]K1 + K1K2 = 0.61 0.21 αA2- = K1K2 [H+]2 + [H+]K1 + K1K2 = 0.23 0.79 pK1 = 3.128 pK2 = 4.761 pK3 = 6.396 Chapter 11: Supplementary Solutions 39 S11-17. K1 = 2.2 × 10-11 K2 = 1 × 10-12 pH: 10.00 10.66 11.00 12.00 12.50 αH2A 0.82 0.49 0.29 0.022 0.003 4 αHA- 0.18 0.49 0.64 0.49 0.24 αA2- 0.001 8 0.022 0.064 0.49 0.76 S11-18. Isoelectric pH = pK1 + pK2 2 = 7.36 Isoionic pH = K1K2(0.010) + K1Kw K1 + (0.010) ⇒ 4.38 × 10-8 M ⇒ pH = 7.36 S11-19. (a) K = e(ΔS°/R) – (ΔH°/RT) ⇒ ΔS° R = -3.23 (±0.53) and ΔH° RT = T 1.44 (±0.15) × 103 Using the value R = 8.314 5 J mol.K gives: ΔS° = [–3.23 (±0.53)][8.314 5] = –27 (±4) J mol.K ΔH° = [–1.44 (±0.15) × 103][8.314 5] = –12 (±1) kJ mol (b) K = [SO3H-] [HSO3-] = e-[3.23 (±0.53) + 1.44 (±0.15) × 103 (1/T)] For T = 298 K, K = e-[3.23 (±0.53) + (1 440 (±150)) / 298] = e-3.23 (±0.53) + 4.83 (±0.50) K = e1.60 (±0.73) = 4.953 (±?) For propagation of uncertainty, we consider K = ex in Table 3-1: eK K = ex ⇒ eK = Kex = (4.953)(±0.73) = ±3.6 ⇒ K = 5.0 (±3.6) CHAPTER 11: SUPPLEMENTARY PROBLEMS 22 S11-1. Write the chemical reactions whose equilibrium constants are Kb1 and Kb2 for the amino acid serine. Find the values of Kb1 and Kb2. S11-2. Consider the diprotic acid H2A with K1 = 1.00 × 10-5 and K2 = 1.00 × 10-9. Find the pH and concentrations of H2A, HA-, and A2- in the following solutions: (a) 0.100 M H2A, (b) 0.100 M NaHA, and (c) 0.100 M Na2A. S11-3. Find the pH of 0.150 M piperazine monohydrochloride, piperazine . HCl. Calculate the concentration of each form of piperazine in this solution. S11-4. Write down, but do not attempt to solve, the exact equations needed to calculate the composition of one liter of solution containing F1 mol of HCl, F2 mol of disodium ascorbate (Na2A, the salt of a weak acid whose two Ka values may be called K1 and K2), and F3 mol of trimethylamine (a weak base, B, whose equilibrium constant should be called Kb). Include activity coefficients wherever appropriate. S11-5. How many mL of 0.423 M KOH should be added to 5.00 g of tartaric acid (2,3-dihydroxybutanedioic acid, FM 150.08) before diluting to 50 mL to give a buffer of (a) pH 3.00 and (b) pH 4.00? S11-6. How many mL of 0.421 M HCl should be added to 50.0 mL of 0.055 5 M disodium malonate (NaO2CCH2CO2Na, FM 148.03, the salt of malonic acid) to adjust the pH to (a) 6.00 and (b) 3.20? S11-7. How many grams of oxalic acid (FM 90.04) should be mixed with 5.00 g of K2C2O4 (FM 166.22) to give a pH of 3.20 when diluted to 250 mL? S11-8. Starting with the fully protonated species, write the stepwise acid dissociation reactions of the amino acids aspartic acid and arginine. Be sure to remove the protons in the correct order. Which species are the neutral molecules that we call aspartic acid and arginine? S11-9. (a) Find the quotient [H2His+]/[HHis] in a 0.050 0 M histidine solution. (b) Find the same quotient for 0.050 0 M histidine monohydrochloride (His . HCl). S11-10. Find the pH and concentration of each species of arginine in 0.012 0 M arginine.HCl solution. S11-11. Consider the neutral base pyrrolidine, C4H9N, designated B. (a) Which is the predominant species, B or BH+, at pH 11? at pH 12? ACID-BASE TITRATIONS Chapter 11: Supplementary Problems 23 (b) At what pH is [BH+] = [B]? (c) What is the quotient [B]/[BH+] at pH 12.00? at pH 2.00? S11-12. Which is the predominant form of sulfurous acid at pH (a) 2, (b) 4, (c) 6 and (d) 8? S11-13. What is the charge of the predominant form of citric acid at pH 5.00? S11-14. The acid HA has pKa = 7.00. Use Equations 11-17 and 11-18 to find the fraction in the form HA and the fraction in the form A- at pH = 8.00. Does your answer agree with what you expect for the quotient [A-]/[HA] at pH 8.00? S11-15. A dibasic compound, B, has pKb1 = 2.00 and pKb2 = 8.00. Find the fraction in the form BH 2 2+ at pH 9.00 using Equation 11-19. Note that K1 and K2 in Equation 11-19 are acid dissociation constants for BH 2 2+ (K1 = Kw/Kb2 and K2 = Kw/Kb1) . S11-16. What fraction of 1,6-hexanedoic acid (adipic acid) is in each form (H2A, HA-, A2-) at pH 5.00? at pH 6.00? S11-17. Calculate αH2A, αHA-, and αA2- for butane-2,3-dione dioxime (dimethylglyoxime) at pH 10.00, 10.66, 11.00, 12.00, and 12.50. S11-18. Calculate the isoelectric and isoionic pH of 0.010 M 8-hydroxyquinoline. S11-19. Thermodynamics and propagation of uncertainty. The bisulfite ion exists in the following equilibrium forms: H S O O O - S OH O O - HSO-3 SO3HThe temperature dependence of the equilibrium constant at an ionic strength of 1.0 M is ln K = –3.23 (±0.53) + 1.44 (±0.15) × 103 × (1/T), where T is temperature in kelvins. Because ln K must be dimensionless, the number –3.23 is dimensionless and the number 1.44 × 103 has the unit of kelvins. (a) Using Equation 6-8, calculate the enthalpy change, ΔH°, and the entropy change, ΔS°, for the isomerization reaction. Include uncertainties in your answers. (b) Calculate the quotient [SO3H-]/[HSO3 - ] at 298 K, including estimated uncertainty.

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