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

3. 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)

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?

5. 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

6. 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)

7. 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.

8. 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

9. The Basic Experimental Arrangement
Syringe
NaHCO3 / NaCl
HCl

10. 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

11. 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

12. 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.

13. 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 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 L
1 mmole occupies 22.4 mL

14. 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 = mmol)
Is this volume significant compared to the ~ 60 mL of CO2 produced?

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

16. 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)
= mL (1 drop ~ mL)
B = Just about 1 drop

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

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

19. 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

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

21. 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

22. 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

23. 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 atm)
2.3% to 4.2% for P ~1 atm

24. 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

25. 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

26. 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 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

27. 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

28. 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 mL
We use 10.0 mL of HCl
P = 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

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

30. 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. %

31. 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%

32. 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

33. 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

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

35. Calculations Posted Measured Calculated Posted vfinal - vinit
Weight of Sample g
Volume of gaseous CO2 v mL
Pressure, P = 752 mm Hg = atm
Temperature, T = 23oC = K
23oC (from Table) 21 mm Hg atm
mmol CO2 (gas) = (P - PH2O)v / RT 1.81 mmol
mmol CO2 (liquid) (Henry’s Law) mmol
Tot CO mmol
Initial = mL
Final = 50.7 mL
752 mm Hg =
752 / 760 atm
= atm
21 mmHg =
21/760
= atm
23oC = K
= 296 K
A more elaborate calculation page is given on the web site.

36. 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 mL
That makes the initial volume,
VI =
= mL
Vsys = mL
Syringe:
Initial = mL
Final = 50.7 mL
and the final volume,
VF =
= mL
PCO2 = ( P - PH2O )( 1 – vI / vF)
= ( – 0.028)( 1 – 95.0 / 140.7)
= atm
SCO2 = 3.2 X 10-2 * atm = M
nCO2s = mL * mmol/mL = 0.10 mmol

37. Calculations Calculated Weight of Sample 0.2147 g
Volume of gaseous CO2 v mL
Pressure, P = 752 mm Hg = atm
Temperature, T = 23oC = K
23oC (from Table) 21 mm Hg atm
mmol CO2 (gas) = (P - PH2O)v / RT 1.81 mmol
mmol CO2 (liquid) (Henry’s Law) mmol
Tot CO mmol
A more elaborate calculation page is given on the web site.
mmol NaHCO3 (from stoichiometry) 1.91 mmol
Weight of NaHCO X g
% NaHCO3 = 100 X / = 74.5 %

38. 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.

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

41. 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)
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