2Gasometric Determination si = kHPiGasometric Determinationof NaHCO3in aMixtureP= Σ Pi(P+a/v2)(v – b) = RT
3CHE 133 MAKE-UP LABORATORY EXERCISE Mon 11/19 or Tue 11/20 Eligible Students are ONLY those who haveexcused absences fromeither a TEST (105 point) or PRELIMINARY (55 point) ExerciseMUST SIGN UP ON SHEETS POSTEDIN EACH LABORATORY ROOMBY November 15(details will follow)
4? 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?
5Determine the percent of NaHCO3 in a mixture by Gasometry Objective:Determine the percent of NaHCO3in a mixture by GasometryConcepts:Ideal Gas Law Henry’s LawVapor Pressure StoichiometryTechniques:Capture Gaseous ProductCorrections to volumeThe students have recently studied the ideal gas law in CHE 131.Apparatus:Gas Syringe Thermometer Barometer
6The Exercise is Conceptually Simple. The unknowns consist of a uniform mixture of NaHCO3 and NaClWeigh Sample, wSample.2. Do Chemistry – Reaction with excess HClNaHCO3 (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)
7to get number of moles of CO2, nCO2 4. Use Ideal Gas Lawto get number of moles of CO2, nCO2P v = n R Tn CO2 = P v CO2 / RTMeasure P, v & T5. From Stoichiometry of reaction, get moles of NaHCO3nNaHCO3 = nCO26. From number of moles, get weightwNaHCO3 = nNaHCO3 * 84.0 g / mol7. Compute Percent Composition of SamplePctNaHCO3 = 100 * wNaHCO3 / wSampleIn 4. PCO2 should really be P – PH2O, but no need to make a big deal of it here, since we derive the required expression later in the lecture.
8But - it Involves Some Important Concepts molesLimiting ReagentsGas Mixtures – Partial PressureGas Solubility – Henry’s LawDoes CO2 behave as an ideal gas?Accuracy & ReproducibilityP= Σ Pisi = kHPi(P+a/v2)(v – b) = nRT
10If we measure volume of CO2 before the sample has reacted completely, the calculated percentage of NaHCO3 will be ______ the actual value.smaller thanequal tolarger than
11Too little CO2 implies too little NaHCO3 and If we measure volume of CO2 before the sample has reacted completely, the calculated percentage of NaHCO3 will be ______ the actual value.1 mol NaHCO3 1 mol CO2Volume 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.A Smaller than
12How 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 HClStoichiometery is 1 : 1.200 mg NaHCO3/(84 mg/mmol ) = 2.4 mmol max of unknown need 2.4 mmol of HCl max to consume itHCl is 1.0 M = 1.0 mmol /mLTherefore, need at most 2.4 mmol/ 1.0 mmol/L = 2.4 mLAre using 10 mL of 1.0 M HCl ( 10 X 1.0 = 10 mmol )– a significant excess Unknown will be the limiting reagentAssume unknown is pure NaHCO3Our assumption is the worst case. The lab manual says to weigh about 200 mg. Even if we used 400 mg, unknown would be limiting.
13On the analytical balance 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 balanceNaHCO3 (s) + H+ (aq) Na+ (aq) + H2O (l) + CO2 (g)1 mol NaHCO3 1 mol CO20.200 g = 200 mg = 2.4 mmolv = n R T/ P = 2.4 X X 298 / 1.0 = 59 mL(Syringe capacity = 60 mL but unknowns 100% NaHCO3)We ask them to weigh only 0.2 g.At STP: mole occupies 22.4 L1 mmole occupies 22.4 mL
14How 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 H2OStill, considering pure NaHCO3,We produce at most 2.4 mmol of liquid H2O(200 mg NaHCO3 = mmol)Is this volume significant compared to the ~ 60 mL of CO2 produced?
15The volume of 2.4 mmol of liquid H2O is about 1 mLabout 1 dropa few mL
16The volume of 2.4 mmol of liquid H2O is: 2. 4 mmol X 18 mg/mmol = 43.2 mgDensity of water = 1 g/mL = 1000 mg/mLV of 43.2 mg = 43.2 mg/ (1000 mg/L)= mL (1 drop ~ mL)B = Just about 1 drop
17What Percent of Gas Volume is the H2O we Produce in the Reaction? Assume we collect ~50 mL of gas100 X mL= %50 mLVolume of water produced by the reaction is < 0.1% of volume of gas collected
18When reaction is complete, all the sodium in the initial sample is present as aqueous NaCl. TrueFalse
19When 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 NaClA = True
20The ApparatusEXTENSION CLAMPRIGHT ANGLE ELBOWSYRINGELARGE TEST TUBESMALL TEST TUBE
21Some of the CO2 will dissolve in the water What’s in the System?SpeciesInitial (mmol)FinalH2O (l)556 mmol556 + x mmolHCl (aq)10 mmol10 – x mmolNaHCO3 (s)x mmol0 mmolNaCl (s)y mmolAir (g)w mmolNaCl (aq)x + y mmolCO2 (g)x - z mmolCO2 (aq)z mmol10 mL of 1.0 M HCl contains:10 X 1.0 = 10 mmol HCl10 mL H2O = 10 g H2O10 g = 10,000 mg / 18.0 mg/mmol= 556 mmol H2ONaHCO3 (s) + H+ (aq) + Cl- (aq) Na+ (aq) + H2O (l) + CO2 (g) + Cl- (aq)Some of the CO2 will dissolve in the water
22Gas Mixtures – Partial Pressure Reaction is conducted in a closed systemat 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) PH2Othe air in the system PairPatmAnd a small amount of non-volatile stuff:200 mg of NaHCO3/NaClPH2O + Pair = PatmWe ignore the contribution of gaseous HCl to the pressure. The Henry’s Law constant for HCl is about 20 M/atm so the initial 1 M solution of HCl has a vapor pressure of about0.05 atm = 38 mm Hg which is smaller than, but comparable with, that of water for which we do correct. Since the HCl is in 4 fold excess or greater, the final solution will have a concentration > 0.75 M or a partial pressure of about 30 mm Hg – a difference of 8 mm Hg. Should we correct for this? I think this would be overkill and it contributes only a little more than 8/760 ~ 0.1% to the error. So we can consider HCl (g) as part of the “air”. (Put a footnote in the written procedure.)We also ignore the volume of the initial solid phase. The densities of both NaHCO3 and NaCl are about 2.2 g/mL so 200 mg will have a volume of less than 0.1 mL – less than the precision of the syringe.And after reaction, PH2O, Pairandthe liberated CO2 PCO2PH2O + Pair + PCO2 = Patm
23What is the Magnitude of PH2O? Table showing vapor pressure of water as a function of temperature is posted in lab.PH2O is a function of temperatureA table of PH2O vs temperature is given in the lecture text. A reminder to students of mm Hg as a measure of pressure. They may actually only have learned about Pascals.Over the range 20oC – 30oC, PH2O increases from:Pressure is often measured in mm of Hg (Torr)1 atm = 760 mm Hg17.5to 31.8 mm Hg(0.023 to atm)2.3% to 4.2% for P ~1 atm
24P,T & nair don’t change, so PiH2O ,T, vi, nairPH2OPfH2O,T, vf, nair, PfCO2PH2OPPnCO2gH2O(l)nCO2sH2O(l)Initially, we haveFinally, we haveP = Piair + PiH2OP = Pfair + PfH2O + PfCO2P,T & nair don’t change, soPiH2O = PfH2O = PH2O andP – PH2O = PiairP – PH2O = Pfair + PfCO2This derives the relationship given in the exercise. It is noted that the bottom line is equation 4 in the exercise.Throughout the exercise, we ignore the HCl in the gas phase. Only the most astute student will realize this. If anyone does, we can do a more exact analysis in private. This analysis is complicated enough for most students.I have written PH2O instead of nH2O in the system based on the previous slide. We approximate that the reaction is conducted both isopiestically and isothermally, so PH20 will be constant even though nH20 is technically not. If we included the change in nH2O, we would need to include the HCl as well.P = n RT/vPiair = Pfair + PfCO2PfCO2 = Piair - PfairnCO2g RT/vftot = nair RT/vitot - nair RT/vftot
25nCO2g RT/vftot = nair RT/vitot - nair RT/vftot nCO2g /vftot = nair /vitot - nair /vftotnCO2g = nair vftot (1/vitot - 1/vftot)But, nair = (P - PH2O) vitot / RTnCO2g = [(P - PH2O) vitot / RT] vftot (1/vitot - 1/vftot)nCO2g = (P - PH2O) (vftot - vitot) / RTEquation 4What is partial pressure of CO2 at end of the reaction?The top line repeats the bottom line on the previous slide. Point out that in Eq 4, the volume difference is just the difference in the initial and final syringe readings. In Equation 6, it the difference is divided by the total final volume! The students should understand this derivation!!!!!! It is just simple algebra.PFCO2 = nCO2g RT/vftotPfCO2 = [(P - PH2O) (vftot - vitot) /RT ] RT/vftotPfCO2 = (P - PH2O) (vftot – vitot)/vftotEquation 6
26What are the various volumes in the exercise? vtube: The volume of just the large test tube and the right angle elbowvtubeYou will measure this!vitot & vftot: The gas phase* volume of the entire closed system before & after the reactionSince the densities of NaCl and NaHCO3 are both close to 2.2 g/mL, the volume of the solid is about g/(2.2 g/mL) ~0.1 mLvtube – vHCl + visyringe*We must exclude the volume of the liquid HCl but can ignore the solid NaHCO3
27vftot – vitot = vtube – vHCl + vfsyringe – vtube + vHCl – visyringe NOTE THAT:Note that:vftot – vitot = vtube – vHCl + vfsyringe – vtube + vHCl – visyringeEquation 4vF – vI = vSYS – vHCl + vfsyringe – vSYS + vHCl – visyringeBUTEquation 61 - vitot / vftotMust emphasize that the total volume is what matters in equation 6 – not the syringe volume.The most common errir in this exercise is to use syringe volumes in equation 6
28Gas Solubility – Henry’s Law nCO2s mol of liberated CO2 dissolves in water.Using Henry’s Law, can calculate concentrationof CO2 dissolved in water.SCO2 = kH * PCO2 ( kH = 3.2 X 10-2 mol / L-atm )kH Given in SUSB-054Suppose:vtube - volume of large test tube & right angle elbow is95.0 mLInitial syringe reading is 5.0 mL and the final reading is mLWe use 10.0 mL of HClP = atm,T = 22oCThe choice of P = 0.32 is justified in the Web supplement to which the students have been directed. An unanswered issue is whether the system really comes to equilibrium by the time the run is over.From the table of PH2O vs TPH2O =20 mm Hg0.026 atm
29PCO2 = ( P - PH2O ) ( 1 – vitot / vftot) Equation 6vitot = mL – 10.0 mL mL = mLvftot = mL – 10.0 mL mL = mLPCO2 = ( – 0.026) ( 1 – 90.0 / 138.0)PCO2 = atmSCO2 = 3.2 X 10-2 mol/L-atm * atm = MWe use 10.0 mL of HClnCO2s = mL * mmol/mL = 0.11 mmol
30What percentage error does 0. 1 mmol represent out of the typical 3 What percentage error does 0.1 mmol represent out of the typical 3.0 mmol of CO2 liberated in this exercise0.33 %3.3 %33. %
31What percentage error does 0. 1 mmol represent out of the typical 3 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%
32Departure 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 ) = RTv1 molFor CO2 at Room Temperature, the corrections to P and v are:a / v2b / vCO20.6%0.18%The van der Waals equation is usually written in terms of the molar volume. I am fuzzing the issue here by writing the n on the right hand side to reduce confusion. The van der Waals constants for CO2 are given in the lecture text – The corrections are calculated as percents of P and v (molar volume) respectively.Small corrections compared to others
33If we seek accuracy to within 1% in the % NaHCO3 ReviewIssueAffectedVariableEffect%Volume of H2O ProducedvCO2g0.1XNon -Ideality of CO2PCO20.6Vapor Pressure of Water2.3 – 4.2Solubility of CO2nCO23.0Now we review which “corrections” that we considered we need to worry about. The 1% accuracy in the % is our arbitrary goal.If we seek accuracy to within 1% in the % NaHCO3
34(Factors affecting Precision) Reproducibility(Factors affecting Precision)wsample analytical balance ( /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%The syringe must be read to nearest 0.5 mL to get 1% precision.Precision should be about 1%Accuracy should be less than 3%* Assuming no ambient temperature changes – see Prob 2
35Calculations Posted Measured Calculated Posted vfinal - vinit Weight of Sample gVolume of gaseous CO2 v mLPressure, P = 752 mm Hg = atmTemperature, T = 23oC = K23oC (from Table) 21 mm Hg atmmmol CO2 (gas) = (P - PH2O)v / RT 1.81 mmolmmol CO2 (liquid) (Henry’s Law) mmolTot CO mmolInitial = mLFinal = 50.7 mL752 mm Hg =752 / 760 atm= atm21 mmHg =21/760= atm23oC =K= 296 KA more elaborate calculation page is given on the web site.
36nCO2s = 10.0 mL * 0.010 mmol/mL = 0.10 mmol To calculate the Henry’s Law correction, we need the Volume of the System. Suppose it is mLThat makes the initial volume,VI == mLVsys = mLSyringe:Initial = mLFinal = 50.7 mLand the final volume,VF == mLPCO2 = ( P - PH2O )( 1 – vI / vF)= ( – 0.028)( 1 – 95.0 / 140.7)= atmSCO2 = 3.2 X 10-2 * atm = MnCO2s = mL * mmol/mL = 0.10 mmol
37Calculations Calculated Weight of Sample 0.2147 g Volume of gaseous CO2 v mLPressure, P = 752 mm Hg = atmTemperature, T = 23oC = K23oC (from Table) 21 mm Hg atmmmol CO2 (gas) = (P - PH2O)v / RT 1.81 mmolmmol CO2 (liquid) (Henry’s Law) mmolTot CO mmolA more elaborate calculation page is given on the web site.mmol NaHCO3 (from stoichiometry) 1.91 mmolWeight of NaHCO X g% NaHCO3 = 100 X / = 74.5 %
38Do test run - then 4 which you report. Procedure - NotesEveryone should weigh ~ 200 mg = 0.2 g– ACCURATELY – for their initial runDepending 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 usingSet syringe at 5.0 mL initially – read to 1 decimal– remember to subtract initial from final volumeDo test run - then 4 which you report.
39Determination of NaHCO3 in a Mixture Last ExerciseDetermination of NaHCO3 in a MixturePart 2 - GravimetricFinal Exercise – 105 pointsDo SUSB Pre-lab Assignment 2Final Quiz will be given at the beginning of the check-out laboratory meeting
41MAKE-UP LABORATORY EXERCISE Monday Dec X at 4:30 PM or CHE 133MAKE-UP LABORATORY EXERCISEMonday Dec X at 4:30 PM orWed Dec X at 4:00 PMEligible Students are ONLY those who have excused absences fromeither a TEST (105 point) or PRELIMINARY (55 point) ExerciseMUST SIGN UP WITH DR. AKHTAR BY Nov XXAll students will do SUSB-055 (Do pre-lab)Download at: