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of NaHCO 3

s i = k H P i. Gasometric Determination. of NaHCO 3. in a. Mixture. P= Σ P i. (P+a/v 2 )(v – b) = RT. CHE 133 MAKE-UP LABORATORY EXERCISE Mon 11/19 - 2:00 PM or Tue 11/20 – 2:30 PM. Eligible Students are ONLY those who have excused absences from

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of NaHCO 3

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  1. si = kHPi Gasometric Determination of NaHCO3 in a Mixture P= ΣPi (P+a/v2)(v – b) = RT

  2. CHE 133MAKE-UP LABORATORY EXERCISE Mon 11/19 - 2:00 PM or Tue 11/20 – 2:30 PM Eligible Students are ONLY those who have excused absences from either a TEST (105 point) orPRELIMINARY (55 point) Exercise MUST SIGN UPON SHEETS POSTED IN EACH LABORATORY ROOM BY November 15 Everyone does the same make-up exercise. You can download the exercise at: http://www.ic.sunysb.edu/Class/che133/susb/SUSB055.pdf

  3. ? QUESTIONS ? How can the composition of a mixture be determined by measuring the volume of gas evolved when that mixture undergoes a chemical reaction? What principles must be considered in using the gas volume as a measure of the amount of gas liberated?

  4. Objective: Determine the percent of NaHCO3 in a mixture by Gasometry Concepts: Ideal Gas Law Henry’s Law Vapor Pressure Stoichiometry Techniques: Capture Gaseous Product Corrections to volume Apparatus: Gas Syringe Thermometer Barometer

  5. The Exercise is Conceptually Simple. • The unknowns consist of a uniform • mixture of NaHCO3 and NaCl • Weigh Sample, wSample. • 2. Do Chemistry – Reaction with excess HCl • NaHCO3(s) + H+ (aq)  Na+ (aq) + H2O (l) + CO2 (g) • NaCl(s)  Na+ (aq) + Cl-(aq) • 3. Capture Liberated Gas and Measure its volume,vCO2 + Cl- (aq) + Cl- (aq) (NaCl will dissolve in, but not react with, HCl)

  6. 4. Use Ideal Gas Law to get number of moles of CO2, nCO2 n CO2= Pv CO2/ RT P v = nR T Measure P, v & T 5. From Stoichiometryof reaction, get moles of NaHCO3 nNaHCO3 = nCO2 6. From number of moles, get weight wNaHCO3 = nNaHCO3 * 84.0 g / mol 7. Compute Percent Composition of Sample PctNaHCO3 = 100 * wNaHCO3 / wSample

  7. But - it Involves Some Important Concepts moles • Limiting Reagents • Gas Mixtures – Partial Pressure • Gas Solubility – Henry’s Law • Does CO2 behave as an ideal gas? • Accuracy & Reproducibility P= ΣPi si = kHPi (P+a/v2)(v – b) = nRT

  8. The Basic Experimental Arrangement Syringe NaHCO3 / NaCl HCl

  9. If we measure volume of CO2 before the sample has reacted completely, the calculated percentageof NaHCO3 will be ______ the actual value. • smaller than • equal to • larger than

  10. If we measure volume of CO2 before the sample has reacted completely, the calculated percentage of NaHCO3will be ______ the actual value. 1 mol NaHCO3 1 mol CO2 Volume of CO2 is proportional to moles (& therefore weight) of NaHCO3 which have reacted. Too little CO2 implies too little NaHCO3 and  too low a percentage. ASmaller than

  11. How Much HCl is needed to insure that Unknown is the Limiting Reagent? NaHCO3 (s) + H+ (aq)  Na+ (aq) + H2O (l) + CO2 (g) 1 mol NaHCO3 1 mol HCl • Stoichiometery is 1 : 1. • 200 mg NaHCO3/(84 mg/mmol ) = 2.4 mmol max of unknown •  need 2.4 mmol of HCl max to consume it • HCl is 1.0 M = 1.0 mmol /mL • Therefore, need at most 2.4 mmol/ 1.0 mmol/L = 2.4 mL • Are using 10 mLof 1.0 M HCl ( 10 X 1.0 = 10 mmol ) • – a significant excess •  Unknown will be the limiting reagent Assume unknown is pure NaHCO3

  12. How Much CO2 is produced? You weigh ~ 0.2 g of unknown. What is maximum volume of CO2 we can expect? ( with P =1.0 atm and T =25oC=298oK) On the analytical balance NaHCO3 (s) + H+ (aq) Na+ (aq) + H2O (l) + CO2 (g) 1 mol NaHCO3 1 mol CO2 0.200 g = 200 mg = 2.4 mmol v = n R T/ P = 2.4 X 0.0821 X 298 / 1.0 = 59mL (Syringe capacity = 60mL but unknowns  100% NaHCO3) At STP: 1 mole occupies 22.4 L 1 mmole occupies 22.4 mL

  13. How much H2O (l) is produced by the reaction? NaHCO3 (s) + H+ (aq)  Na+ (aq) + H2O (l) + CO2 (g) Reaction produces water. 1 mol NaHCO3 1 mol H2O Still, considering pure NaHCO3, We produce at most 2.4 mmol of liquid H2O (200 mg NaHCO3 = 2.4 mmol) Is this volume significant compared to the ~ 60 mL of CO2 produced?

  14. The volume of 2.4 mmol of liquid H2O is • about 1 mL • about 1 drop • a few mL

  15. The volume of 2.4 mmol of liquid H2O is: 2. 4 mmol X 18 mg/mmol = 43.2 mg Density of water = 1 g/mL = 1000 mg/mL V of 43.2 mg = 43.2 mg/ (1000 mg/L) = 0.0432 mL (1 drop ~ 0.050 mL) B= Just about 1 drop

  16. What Percent of Gas Volume is the H2O we Produce in the Reaction? Assume we collect ~50 mL of gas 100 X 0.042 mL --------------------- = 0.08 % 50 mL Volume of water produced by the reaction is < 0.1% of volume of gas collected

  17. When reaction is complete, all the sodium in the initial sample is present as aqueous NaCl. • True • False

  18. When reaction is complete, all the sodium in the initial sample is present as aqueous NaCl. NaHCO3 (s) + H+ (aq)  Na+ (aq) + H2O (l) + CO2 (g) Cl- Cl- The samples consist of NaHCO3 and NaCl. In the reaction with HCl, all of the NaHCO3 is converted to NaCl A= True

  19. The Apparatus EXTENSION CLAMP RIGHT ANGLE ELBOW SYRINGE LARGE TEST TUBE SMALL TEST TUBE

  20. What’s in the System? 10 mL of 1.0 M HCl contains: 10 X 1.0 = 10 mmol HCl 10 mL H2O = 10 g H2O 10 g = 10,000 mg / 18.0 mg/mmol = 556 mmol H2O NaHCO3 (s) + H+ (aq) + Cl- (aq)  Na+ (aq) + H2O (l) + CO2 (g) + Cl- (aq) Some of the CO2 will dissolve in the water

  21. Gas Mixtures – Partial Pressure • Reaction is conducted in a closed system • at constant external pressure • - atmospheric (P ~ 1 atm) • at constant temperature • - room temperature (T ~ 25oC) • Initially: • System contains air & water (HCl) • Pressure in system is potentially due to: • water (from HCl)PH2O • the air in the system Pair Patm And a small amount of non-volatile stuff: 200 mg of NaHCO3/NaCl PH2O + Pair = Patm And after reaction, PH2O, Pair • and • the liberated CO2PCO2 PH2O + Pair + PCO2 = Patm

  22. What is the Magnitude of PH2O? Table showing vapor pressure of water as a function of temperature is posted in lab. PH2Ois a function of temperature Over the range 20oC – 30oC, PH2Oincreases from: Pressure is often measured in mm of Hg (Torr) 1 atm = 760 mm Hg to 31.8 mm Hg 17.5 (0.023 to 0.042 atm) 2.3% to 4.2% for P ~1 atm

  23. PH2O PiH2O ,T, vi, nair PH2O PfH2O,T, vf, nair, PfCO2 P P nCO2g nCO2s H2O(l) H2O(l) Initially, we have Finally, we have P = Piair + PiH2O P = Pfair + PfH2O + PfCO2 P,T& nair don’t change, so PiH2O = PfH2O = PH2O and P – PH2O = Piair P – PH2O = Pfair+ PfCO2 Piair = Pfair+ PfCO2 P = nRT/v PfCO2 = Piair- Pfair nCO2g RT/vftot= nair RT/vitot - nair RT/vftot

  24. nCO2g RT/vftot= nair RT/vitot- nair RT/vftot nCO2g /vftot= nair /vitot- nair /vftot nCO2g= nair vftot(1/vitot-1/vftot) But, nair = (P - PH2O) vitot/ RT nCO2g= [(P - PH2O) vitot/ RT]vftot(1/vitot - 1/vftot) nCO2g= (P - PH2O) (vftot- vitot) / RT Equation 4 What is partial pressure of CO2 at end of the reaction? PFCO2 = nCO2g RT/vftot PfCO2 = [(P - PH2O) (vftot- vitot) /RT ] RT/vftot PfCO2 = (P - PH2O) (vftot– vitot)/vftot Equation 6

  25. What are the various volumes in the exercise? vtube: The volume of just the large test tube and the right angle elbow vtube You will measure this! vitot & vftot: The gas phase* volume of the entire closed system before & after the reaction vtube–vHCl+visyringe *We must exclude the volume of the liquid HCl but can ignore the solid NaHCO3

  26. NOTE THAT: Note that: vftot – vitot = vtube – vHCl + vfsyringe – vtube + vHCl – visyringe Equation 4 vF – vI = vSYS – vHCl + vfsyringe– vSYS + vHCl – visyringe BUT Equation 6 1 - vitot / vftot The most common errir in this exercise is to use syringe volumes in equation 6

  27. Gas Solubility – Henry’s Law nCO2s mol of liberated CO2 dissolves in water. Using Henry’s Law, can calculate concentration of CO2dissolved in water. SCO2 = kH * PCO2 ( kH = 3.2 X 10-2 mol / L-atm) kH Given in SUSB-054 • Suppose: • vtube - volume of large test tube & right angle elbow is • 95.0 mL • Initial syringe reading is 5.0 mL and the final reading is 53.0 mL • We use 10.0 mL of HCl • P = 1.000 atm, • T = 22oC From the table of PH2O vs T PH2O = 20 mm Hg 0.026 atm

  28. PCO2 = ( P - PH2O) ( 1 – vitot / vftot) Equation 6 vitot = 95.0 mL – 10.0 mL + 5.0 mL = 90.0 mL vftot = 95.0 mL – 10.0 mL + 53.0 mL = 138.0 mL PCO2 = ( 1.000 – 0.026) ( 1 – 90.0 / 138.0) PCO2 = 0.339 atm SCO2 = 3.2 X 10-2 mol/L-atm * 0.339 atm = 0.011 M We use 10.0 mL of HCl nCO2s = 10.0 mL * 0.011 mmol/mL = 0.11 mmol

  29. What percentage error does 0.1 mmol represent out of the typical 3.0 mmol of CO2 liberated in this exercise • 0.33 % • 3.3 % • 33. %

  30. What percentage error does 0.1 mmol represent out of the typical 3.0 mmol of CO2 liberated in this exercise. 100 X 0.1 / 3.0 = 3.3 % B= 3.3%

  31. Departure of gases from Ideality? Can CO2 be treated as an Ideal Gas? We can estimate the deviation from ideality by examining the van der Waals constants. ( P + a / v2 ) ( v – b) = RT v 1 mol For CO2 at Room Temperature, the corrections to P and v are: Small corrections compared to others

  32. Review If we seek accuracy to within 1% in the % NaHCO3

  33. Reproducibility (Factors affecting Precision) wsample analytical balance (0.0002 /0.2000) ~0.1% PCO2 barometer (1 / 760) ~0.1% PH2O table of values (1 / 760) ~0.1% T thermometer (1/ 300)* ~0.3% vCO2g syringe (0.5 / 50) ~1% Precision should be about 1% Accuracy should be less than 3% * Assuming no ambient temperature changes – see Prob 2

  34. Calculations vfinal - vinit Posted Measured Posted Calculated Weight of Sample0.2147 g Volume of gaseous CO2v45.7 mL Pressure, P = 752 mm Hg = 0.989 atm Temperature, T = 23oC = 296 K PH2O@ 23oC (from Table) 21 mm Hg 0.028 atm mmol CO2 (gas) = (P - PH2O)v / RT 1.81 mmol mmol CO2 (liquid) (Henry’s Law) 0.10 mmol Tot CO2 1.91 mmol Initial = 5.0 mL Final = 50.7 mL 752 mm Hg = 752 / 760 atm = 0.989 atm 23oC = 23 + 273 K = 296 K 21 mmHg = 21/760 = 0.028 atm

  35. To calculate the Henry’s Law correction, we need the total volume of the system. Suppose vtube is 100.0 mL • That makes the initial volume, • vitot = 100.0 - 10.0 + 5.0 • = 95.0 mL vtube = 100.0 mL Syringe: Initial = 5.0 mL Final = 50.7 mL • and the final volume, • vftot = 100.0 - 10.0 + 50.7 • = 140.7 mL PCO2 = ( P - PH2O)( 1 – vI / vF) = ( 0.989 – 0.028)( 1 – 95.0 / 140.7) = 0.312 atm SCO2 = 3.2 X 10-2* 0.312 atm = 0.010 M nCO2s = 10.0 mL * 0.010 mmol/mL = 0.10 mmol

  36. Calculated Calculations Weight of Sample0.2147 g Volume of gaseous CO2v45.7 mL Pressure, P = 752 mm Hg = 0.989 atm Temperature, T = 23oC = 296 K PH2O@ 23oC (from Table) 21 mm Hg 0.028 atm mmol CO2 (gas) = (P - PH2O)v / RT 1.81 mmol mmol CO2 (liquid) (Henry’s Law) 0.10 mmol Tot CO2 1.91 mmol mmol NaHCO3 (from stoichiometry) 1.91 mmol Weight of NaHCO3 1.91 X 84.0 0.160 g % NaHCO3 = 100 X 0.160 / 0.2147 = 74.5 %

  37. Procedure - Notes • Everyone should weigh ~ 200 mg = 0.2 g • – ACCURATELY – for their initial run • Depending on your sample, you may need to adjust the weight in subsequent runs to insure that you get between 30 and 50 mL of CO2, but not more than 50 mL. • Test that system is air-tight before using • Set syringe at 5.0 mL initially – read to 1 decimal • – remember to subtract initial from final volume Do test run - then 4 which you report.

  38. CHE 133MAKE-UP LABORATORY EXERCISE Mon 11/19 - 2:00 PM or Tue 11/20 – 2:30 PM Eligible Students are ONLY those who have excused absences from either a TEST (105 point) orPRELIMINARY (55 point) Exercise MUST SIGN UPON SHEETS POSTED IN EACH LABORATORY ROOM BY November 15 Everyone does the same make-up exercise. You can download the exercise at: http://www.ic.sunysb.edu/Class/che133/susb/SUSB055.pdf

  39. Last Exercise Determination of NaHCO3 in a Mixture Final Exercise – 105 points Part 2 - Gravimetric Do SUSB-054 - Pre-lab Assignment 2 Final Quiz will be given at the beginning of the check-out laboratory meeting

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