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Slide 2 1 Slide 3 of NaHCO 3 Mixture Gasometric Determination in a P= P i s i = k H P i (P+a/v 2 )(v b) = RT 2 Slide 4 3 CHE 133CHE 133 MAKE-UP LABORATORY EXERCISE Mon 11/19 or Tue 11/20 Slide 5 4 ? QUESTIONS ?? QUESTIONS ? Slide 6 Objective: percent of NaHCO 3 Determine the percent of NaHCO 3 in a mixture by Gasometry Concepts: Ideal Gas LawHenrys Law Vapor PressureStoichiometry Techniques: Capture Gaseous Product Corrections to volume Apparatus: Gas SyringeThermometerBarometer 5 Slide 7 The Exercise is Conceptually Simple. The unknowns consist of a uniform mixture of NaHCO 3 and NaCl w Sample 1.Weigh Sample, w Sample. 2. Do Chemistry Reaction with excess HCl NaHCO 3 (s) + H + (aq) Na + (aq) + H 2 O (l) + CO 2 (g) NaCl(s) Na + (aq) + Cl - (aq) v CO2 3. Capture Liberated Gas and Measure its volume, v CO2 6 Slide 8 5. From Stoichiometry of reaction, get moles of NaHCO 3 n NaHCO3 n CO2 n NaHCO3 = n CO2 6. From number of moles, get weight w NaHCO3 n NaHCO3 w NaHCO3 = n NaHCO3 * 84.0 g / mol 7. Compute Percent Composition of Sample w NaHCO3 w Sample Pct NaHCO3 = 100 * w NaHCO3 / w Sample 4. Use Ideal Gas Law P v = n R T to get number of moles of CO 2, n CO2 v CO2 n CO2 = P v CO2 / RT 7 Slide 9 But - it Involves Some Important Concepts Limiting Reagents Gas Mixtures Partial Pressure Gas Solubility Henrys Law Does CO 2 behave as an ideal gas? Accuracy & Reproducibility P= P i s i = k H P i (P+a/v 2 )(v b) = nRT moles 8 Slide 10 NaHCO 3 / NaCl HCl Syringe 9 The Basic Experimental ArrangementThe Basic Experimental Arrangement Slide 11 ______ If we measure volume of CO 2 before the sample has reacted completely, the calculated percentage of NaHCO 3 will be ______ the actual value. 10 A.smaller than B.equal to C.larger than Slide 12 ______ If we measure volume of CO 2 before the sample has reacted completely, the calculated percentage of NaHCO 3 will be ______ the actual value. A Smaller than 11 1 mol NaHCO 3 1 mol CO 2 Volume of CO 2 is proportional to moles (& therefore weight) of NaHCO 3 which have reacted. Too little CO 2 implies too little NaHCO 3 and too low a percentage. Slide 13 Stoichiometery is 1 : 1. 200 mg NaHCO 3 /(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 10 mL Are using 10 mL of 1.0 M HCl ( 10 X 1.0 = 10 mmol ) a significant excess Unknown will be the limiting reagent NaHCO 3 (s) + H + (aq) Na + (aq) + H 2 O (l) + CO 2 (g) 1 mol NaHCO 3 1 mol HCl 12 How Much HCl is needed to insure that Unknown is the Limiting Reagent? Assume unknown is pure NaHCO 3 Slide 14 You weigh ~ 0.2 g of unknown. What is maximum volume of CO 2 we can expect? ( with P = 1.0 atm and T = 25 o C = 298 o K ) NaHCO 3 (s) + H + (aq) Na + (aq) + H 2 O (l) + CO 2 (g) 0.200 g = 200 mg = 2.4 mmol v = n R T/ P = 2.4 X 0.0821 X 298 / 1.0 = 59 mL (Syringe capacity = 60 mL but unknowns 100% NaHCO 3 ) 1 mol NaHCO 3 1 mol CO 2 On the analytical balance 13 How Much CO 2 is produced?How Much CO 2 is produced? Slide 15 Still, considering pure NaHCO 3, 2.4 mmol We produce at most 2.4 mmol of liquid H 2 O NaHCO 3 (s) + H + (aq) Na + (aq) + H 2 O (l) + CO 2 (g) 1 mol NaHCO 3 1 mol H 2 O 14 Is this volume significant compared to the ~ 60 mL of CO 2 produced? Reaction produces water. How much H 2 O (l) is produced by the reaction?How much H 2 O (l) is produced by the reaction? Slide 16 The volume of 2.4 mmol of liquid H 2 O is 15 A.about 1 mL B.about 1 drop C.a few mL Slide 17 16 B = Just about 1 drop 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) The volume of 2.4 mmol of liquid H 2 O is: Slide 18 100 X 0.042 mL --------------------- = 0.08 % 50 mL Volume of water produced by the reaction is < 0.1% of volume of gas collected 17 What Percent of Gas Volume is the H 2 O we Produce in the Reaction? Assume we collect ~50 mL of gas Slide 19 When reaction is complete, all the sodium in the initial sample is present as aqueous NaCl. 18 A.True B.False Slide 20 A = True The samples consist of NaHCO 3 and NaCl. In the reaction with HCl, all of the NaHCO 3 is converted to NaCl NaHCO 3 (s) + H + (aq) Na + (aq) + H 2 O (l) + CO 2 (g) Cl - Cl- 19 When reaction is complete, all the sodium in the initial sample is present as aqueous NaCl. Slide 21 SYRINGE LARGE TEST TUBE RIGHT ANGLE ELBOW SMALL TEST TUBE EXTENSION CLAMP 20 The ApparatusThe Apparatus Slide 22 SpeciesInitial (mmol)Final H 2 O (l)556 mmol556 + x mmol HCl (aq)10 mmol10 x mmol NaHCO 3 (s)x mmol0 mmol NaCl (s)y mmol0 mmol Air (g)w mmol NaCl (aq)0 mmolx + y mmol CO 2 (g)0 mmolx - z mmol CO 2 (aq)0 mmolz mmol 21 10 mL of 1.0 M HCl contains: 10 mmol HCl 10 X 1.0 = 10 mmol HCl 10 mL H 2 O = 10 g H 2 O 10 g = 10,000 mg / 18.0 mg/mmol 556 mmol H 2 O = 556 mmol H 2 O Some of the CO 2 will dissolve in the water NaHCO 3 (s) + H + (aq) + Cl - (aq) Na + (aq) + H 2 O (l) + CO 2 (g) + Cl - (aq) Whats in the System? Slide 23 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 ~ 25 o C) Initially: water System contains air & water (HCl) Pressure in system is potentially due to: P H2O water(from HCl)P H2O the air in the systemP air P atm P H2O And after reaction, P H2O, P air and the liberated CO 2 P CO2 And a small amount of non-volatile stuff: 200 mg of NaHCO 3 /NaCl 22 P H2O P H2O + P air = P atm P H2O P H2O + P air + P CO2 = P atm Slide 24 P H2O Over the range 20 o C 30 o C, P H2O increases from: P H2O P H2O is a function of temperature Pressure is often measured in mm of Hg (Torr) 1 atm = 760 mm Hg 17.5 to 31.8 mm Hg 2.3% to 4.2% for P ~1 atm vapor pressure of water function of temperature Table showing vapor pressure of water as a function of temperature is posted in lab. 23 (0.023 to 0.042 atm) What is the Magnitude of P H2O ?What is the Magnitude of P H2O ? Slide 25 Initially, we have Finally, we have P i H2O,T, v i, n air n CO2 g n CO2 s PP P f H2O,T, v f, n air, P f CO2 H 2 O(l) 24 P P,T & n air dont change, so P H2O P i H2O = P f H2O = P H2O and n CO2 g RT/v f tot = n air RT/v i tot - n air RT/v f tot P H2O P = n RT/v Slide 26 25 n CO2 g RT/v f tot = n air RT/v i tot - n air RT/v f tot n CO2 g /v f tot = n air /v i tot - n air /v f tot n CO2 g = n air v f tot (1/v i tot - 1/v f tot ) PP H2O But, n air = (P - P H2O ) v i tot / RT PP H2O n CO2 g = [(P - P H2O ) v i tot / RT] v f tot (1/v i tot - 1/v f tot ) PP H2O n CO2 g = (P - P H2O ) (v f tot - v i tot ) / RT Equation 4 P F CO2 = n CO2 g RT/v f tot PP H2O P f CO2 = [(P - P H2O ) (v f tot - v i tot ) /RT ] RT/v f tot What is partial pressure of CO 2 at end of the reaction? PP H2O P f CO2 = (P - P H2O ) (v f tot v i tot )/v f tot Equation 6 Slide 27 v tube : The volume of just the large test tube and the right angle elbow v i tot & v f tot : The gas phase* volume of the entire closed system before & after the reaction v tube v HCl + v i syringe v tube 26 What are the various volumes in the exercise?What are the various volumes in the exercise? Slide 28 Note that: v f tot v i tot = v tube v HCl + v f syringe v tube + v HCl v i syringe v F v I = v SYS v HCl + v f syringe v SYS + v HCl v i syringe 1 - v i tot / v f tot BUT NOTE THAT: 27 Slide 29 Gas Solubility Henrys Law n CO2 s mol of liberated CO 2 dissolves in water. Using Henrys Law, can calculate concentration of CO 2 dissolved in water. S CO2 = k H * P CO2 ( k H = 3.2 X 10 -2 mol / L-atm ) Suppose: v tube - 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 = 22 o C k H Given in SUSB-054 28 Slide 30 P CO2 = ( P - P H2O ) ( 1 v i tot / v f tot ) v i tot = 95.0 mL 10.0 mL + 5.0 mL = 90.0 mL v f tot = 95.0 mL 10.0 mL + 53.0 mL = 138.0 mL P CO2 = ( 1.000 0.026) ( 1 90.0 / 138.0) P CO2 = 0.339 atm S CO2 = 3.2 X 10 -2 mol/L-atm * 0.339 atm = 0.011 M We use 10.0 mL of HCl n CO2 s n CO2 s = 10.0 mL * 0.011 mmol/mL = 0.11 mmol Equation 6 29 Slide 31 What percentage error does 0.1 mmol represent out of the typical 3.0 mmol of CO 2 liberated in this exercise 30 A. 0.33 % B. 3.3 % C. 33. % Slide 32 What percentage error does 0.1 mmol represent out of the typical 3.0 mmol of CO 2 liberated in this exercise. B = 3.3% 100 X 0.1 / 3.0 = 3.3 % 31 Slide 33 ( P + a / v 2 ) ( v b ) = RT a / v 2 b / v CO 2 0.6%0.18% For CO 2 at Room Temperature, the corrections to P and v are: Small corrections compared to others Can CO 2 be treated as an Ideal Gas? We can estimate the deviation from ideality by examining the van der Waals constants. 1 molv 32 Departure of gases from Ideality?Departure of gases from Ideality? Slide 34 33 Review Issue Affected Variable Effect % Volume of H 2 O Producedv CO2 g 0.1 X Non -Ideality of CO 2 P CO2 0.6 X Vapor Pressure of WaterP CO2 2.3 4.2 Solubility of CO 2 n CO2 3.0 If we seek accuracy to within 1% in the % NaHCO 3 Slide 35 w sample analytical balance (0.0002 /0.2000) ~0.1% P CO2 barometer(1 / 760) ~0.1% P H2O table of values(1 / 760) ~0.1% T thermometer(1/ 300)* ~0.3% v CO2 g syringe(0.5 / 50) ~1% Precision should be about 1% Accuracy should be less than 3% Reproducibility (Factors affecting Precision) * Assuming no ambient temperature changes see Prob 2 34 Slide 36 Weight of Sample0.2147 g Volume of gaseous CO 2 v45.7 mL Pressure, P = 752 mm Hg = 0.989 atm Temperature, T = 23 o C =296 K P H2O @ 23 o C (from Table) 21 mm Hg0.028 atm mmol CO 2 (gas)= (P - P H2O )v / RT1.81 mmol mmol CO 2 (liquid) (Henrys Law)0.10 mmol Tot CO 2 1.91 mmol v final - v init PostedMeasuredPostedCalculated 752 mm Hg = 752 / 760 atm = 0.989 atm 23 o C = 23 + 273 K = 296 K 21 mmHg = 21/760 = 0.028 atm Initial = 5.0 mL Final = 50.7 mL 35 Calculations Slide 37 To calculate the Henrys Law correction, we need the Volume of the System. Suppose it is 100.0 mL That makes the initial volume, V I = 100.0 -10.0 + 5.0 = 95.0 mL P CO2 = ( P - P H2O )( 1 v I / v F ) = ( 0.989 0.028)( 1 95.0 / 140.7)= 0.312 atm S CO2 = 3.2 X 10 -2 * 0.312 atm = 0.010 M n CO2 s n CO2 s = 10.0 mL * 0.010 mmol/mL = 0.10 mmol and the final volume, V F = 100.0 -10.0 + 50.7 = 140.7 mL V sys = 100.0 mL Syringe: Initial = 5.0 mL Final = 50.7 mL 36 Slide 38 Calculations Weight of Sample0.2147 g Volume of gaseous CO 2 v45.7 mL Pressure, P = 752 mm Hg = 0.989 atm Temperature, T = 23 o C =296 K P H2O @ 23 o C (from Table) 21 mm Hg0.028 atm mmol CO 2 (gas)= (P - P H2O )v / RT1.81 mmol mmol CO 2 (liquid) (Henrys Law)0.10 mmol Tot CO 2 1.91 mmol mmol NaHCO 3 (from stoichiometry)1.91 mmol Weight of NaHCO 3 1.91 X 84.00.160 g % NaHCO 3 = 100 X 0.160 / 0.2147 =74.5 % Calculated 37 Slide 39 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 CO 2, 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 Procedure - NotesProcedure - Notes Slide 40 Determination of NaHCO 3 in a Mixture Part 2 - Gravimetric Final Exercise 105 points Last Exercise Do SUSB-054 - Pre-lab Assignment 2 Final Quiz will be given at the beginning of the check-out laboratory meeting 39 Slide 41 40 Slide 42 41