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

CHE 133 MAKE-UP LABORATORY EXERCISE Mon 11/19 or Tue 11/20 Eligible Students are ONLY those who have excused absences from either a TEST (105 point) or PRELIMINARY (55 point) Exercise MUST SIGN UP ON SHEETS POSTED IN EACH LABORATORY ROOM BY November 15 (details will follow)

? 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?

Determine the percent of NaHCO3 in a mixture by Gasometry 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 The students have recently studied the ideal gas law in CHE 131. Apparatus: Gas Syringe Thermometer Barometer

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)

to get number of moles of CO2, nCO2 4. Use Ideal Gas Law to get number of moles of CO2, nCO2 P v = n R T n CO2 = P v CO2 / RT Measure P, v & T 5. From Stoichiometry of 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 In 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.

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= Σ Pi si = kHPi (P+a/v2)(v – b) = nRT

The Basic Experimental Arrangement Syringe NaHCO3 / NaCl HCl

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

Too 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 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. A Smaller than

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 mL of 1.0 M HCl ( 10 X 1.0 = 10 mmol ) – a significant excess  Unknown will be the limiting reagent Assume unknown is pure NaHCO3 Our assumption is the worst case. The lab manual says to weigh about 200 mg. Even if we used 400 mg, unknown would be limiting.

On 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 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 = 59 mL (Syringe capacity = 60 mL but unknowns  100% NaHCO3) We ask them to weigh only 0.2 g. At STP: 1 mole occupies 22.4 L 1 mmole occupies 22.4 mL

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?

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

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

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

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

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

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

Some of the CO2 will dissolve in the water What’s in the System? Species Initial (mmol) Final H2O (l) 556 mmol 556 + x mmol HCl (aq) 10 mmol 10 – x mmol NaHCO3 (s) x mmol 0 mmol NaCl (s) y mmol Air (g) w mmol NaCl (aq) x + y mmol CO2 (g) x - z mmol CO2 (aq) z mmol 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

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 We 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 about 0.05 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, Pair and the liberated CO2 PCO2 PH2O + Pair + PCO2 = Patm

What 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 temperature A 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 Hg 17.5 to 31.8 mm Hg (0.023 to 0.042 atm) 2.3% to 4.2% for P ~1 atm

P,T & nair don’t change, so PiH2O ,T, vi, nair PH2O PfH2O,T, vf, nair, PfCO2 PH2O P P nCO2g H2O(l) nCO2s 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 This 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/v Piair = Pfair + PfCO2 PfCO2 = Piair - Pfair nCO2g RT/vftot = nair RT/vitot - nair RT/vftot

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? 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/vftot PfCO2 = [(P - PH2O) (vftot - vitot) /RT ] RT/vftot PfCO2 = (P - PH2O) (vftot – vitot)/vftot Equation 6

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 Since the densities of NaCl and NaHCO3 are both close to 2.2 g/mL, the volume of the solid is about 0.200 g/(2.2 g/mL) ~0.1 mL vtube – vHCl + visyringe *We must exclude the volume of the liquid HCl but can ignore the solid NaHCO3

vftot – vitot = vtube – vHCl + vfsyringe – vtube + vHCl – visyringe 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 Must 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

Gas Solubility – Henry’s Law nCO2s mol of liberated CO2 dissolves in water. Using Henry’s Law, can calculate concentration of CO2 dissolved 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 The 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 T PH2O = 20 mm Hg 0.026 atm

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

What 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 0.33 % 3.3 % 33. %

What 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%

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: a / v2 b / v CO2 0.6% 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

If we seek accuracy to within 1% in the % NaHCO3 Review Issue Affected Variable Effect % Volume of H2O Produced vCO2g 0.1 X Non -Ideality of CO2 PCO2 0.6 Vapor Pressure of Water 2.3 – 4.2  Solubility of CO2 nCO2 3.0 Now 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

(Factors affecting Precision) 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% 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

Calculations Posted Measured Calculated Posted vfinal - vinit Weight of Sample 0.2147 g Volume of gaseous CO2 v 45.7 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 21 mmHg = 21/760 = 0.028 atm 23oC = 23 + 273 K = 296 K A more elaborate calculation page is given on the web site.

nCO2s = 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 100.0 mL That makes the initial volume, VI = 100.0 -10.0 + 5.0 = 95.0 mL Vsys = 100.0 mL Syringe: Initial = 5.0 mL Final = 50.7 mL and the final volume, VF = 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

Calculations Calculated Weight of Sample 0.2147 g Volume of gaseous CO2 v 45.7 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 A more elaborate calculation page is given on the web site. 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 %

Do test run - then 4 which you report. 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.

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

MAKE-UP LABORATORY EXERCISE Monday Dec X at 4:30 PM or CHE 133 MAKE-UP LABORATORY EXERCISE Monday Dec X at 4:30 PM or Wed Dec X at 4:00 PM Eligible Students are ONLY those who have excused absences from either a TEST (105 point) or PRELIMINARY (55 point) Exercise MUST SIGN UP WITH DR. AKHTAR BY Nov XX All students will do SUSB-055 (Do pre-lab) Download at: http://www.ic.sunysb.edu/Class/che133/susb/SUSB055.pdf