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1 of NaHCO 3 Mixture Gasometric Determination in a P= Σ P i s i = k H P i (P+a/v 2 )(v – b) = RT 2.

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Presentation on theme: "1 of NaHCO 3 Mixture Gasometric Determination in a P= Σ P i s i = k H P i (P+a/v 2 )(v – b) = RT 2."— Presentation transcript:

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3 of NaHCO 3 Mixture Gasometric Determination in a P= Σ P i s i = k H P i (P+a/v 2 )(v – b) = RT 2

4 3 CHE 133CHE 133 MAKE-UP LABORATORY EXERCISE Mon 11/19 or Tue 11/20

5 4 ? QUESTIONS ?? QUESTIONS ?

6 Objective: percent of NaHCO 3 Determine the percent of NaHCO 3 in a mixture by Gasometry Concepts: Ideal Gas LawHenry’s Law Vapor PressureStoichiometry Techniques: Capture Gaseous Product Corrections to volume Apparatus: Gas SyringeThermometerBarometer 5

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

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

9 But - it Involves Some Important Concepts Limiting Reagents Gas Mixtures – Partial Pressure Gas Solubility – Henry’s Law Does CO 2 behave as an ideal gas? Accuracy & Reproducibility P= Σ P i s i = k H P i (P+a/v 2 )(v – b) = nRT moles 8

10 NaHCO 3 / NaCl HCl Syringe 9 The Basic Experimental ArrangementThe Basic Experimental Arrangement

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

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.

13 Stoichiometery is 1 : 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

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) g = 200 mg = 2.4 mmol v = n R T/ P = 2.4 X 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?

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?

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

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) = mL (1 drop ~ mL) The volume of 2.4 mmol of liquid H 2 O is:

18 100 X 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

19 When reaction is complete, all the sodium in the initial sample is present as aqueous NaCl. 18 A.True B.False

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.

21 SYRINGE LARGE TEST TUBE RIGHT ANGLE ELBOW SMALL TEST TUBE EXTENSION CLAMP 20 The ApparatusThe Apparatus

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 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) What’s in the System?

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

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 atm) What is the Magnitude of P H2O ?What is the Magnitude of P H2O ?

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 don’t 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

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

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?

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

29 Gas Solubility – Henry’s Law n CO2 s mol of liberated CO 2 dissolves in water. Using Henry’s Law, can calculate concentration of CO 2 dissolved in water. S CO2 = k H * P CO2 ( k H = 3.2 X 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 = atm, T = 22 o C k H Given in SUSB

30 P CO2 = ( P - P H2O ) ( 1 – v i tot / v f tot ) v i tot = 95.0 mL – 10.0 mL mL = 90.0 mL v f tot = 95.0 mL – 10.0 mL mL = mL P CO2 = ( – 0.026) ( 1 – 90.0 / 138.0) P CO2 = atm S CO2 = 3.2 X mol/L-atm * atm = M We use 10.0 mL of HCl n CO2 s n CO2 s = 10.0 mL * mmol/mL = 0.11 mmol Equation 6 29

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 % B. 3.3 % C. 33. %

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

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?

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

35 w sample analytical balance ( /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

36 Weight of Sample g Volume of gaseous CO 2 v45.7 mL Pressure, P = 752 mm Hg = atm Temperature, T = 23 o C =296 K P 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) (Henry’s Law)0.10 mmol Tot CO mmol v final - v init PostedMeasuredPostedCalculated 752 mm Hg = 752 / 760 atm = atm 23 o C = K = 296 K 21 mmHg = 21/760 = atm Initial = 5.0 mL Final = 50.7 mL 35 Calculations

37 To calculate the Henry’s Law correction, we need the Volume of the System. Suppose it is mL That makes the initial volume, V I = = 95.0 mL P CO2 = ( P - P H2O )( 1 – v I / v F ) = ( – 0.028)( 1 – 95.0 / 140.7)= atm S CO2 = 3.2 X * atm = M n CO2 s n CO2 s = 10.0 mL * mmol/mL = 0.10 mmol and the final volume, V F = = mL V sys = mL Syringe: Initial = 5.0 mL Final = 50.7 mL 36

38 Calculations Weight of Sample g Volume of gaseous CO 2 v45.7 mL Pressure, P = 752 mm Hg = atm Temperature, T = 23 o C =296 K P 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) (Henry’s Law)0.10 mmol Tot CO mmol mmol NaHCO 3 (from stoichiometry)1.91 mmol Weight of NaHCO X g % NaHCO 3 = 100 X / =74.5 % Calculated 37

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

40 Determination of NaHCO 3 in a Mixture Part 2 - Gravimetric Final Exercise – 105 points Last Exercise Do SUSB Pre-lab Assignment 2 Final Quiz will be given at the beginning of the check-out laboratory meeting 39

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