 # Determination of NaHCO3 in a Mixture

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Gasometric Determination
of NaHCO3
in a
Mixture
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CHE 133
MAKE-UP LABORATORY EXERCISE
Mon 11/19 - 2:00 PM or Tue 11/20 – 2:30 PM
Eligible Students are ONLY those who have
excused absences from
either a TEST (105 point)
or
PRELIMINARY (55 point) Exercise
IN EACH LABORATORY ROOM
BY November 15
Everyone does the same make-up exercise. You can download the
exercise at:
http://www.ic.sunysb.edu/Class/che133/susb/SUSB055.pdf 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
Objective:
Determine the percent of NaHCO3
in a mixture by Gasometry
Concepts:
Ideal Gas Law
Vapor Pressure
Henry’s Law
Stoichiometry
Techniques:
Capture Gaseous Product
Corrections to volume
Apparatus:
Gas Syringe
Thermometer
Barometer
6
The Exercise is Conceptually Simple.
The unknowns consist of a uniform
mixture of NaHCO3 and NaCl
1. Weigh Sample, wSample.
2. Do Chemistry – Reaction with excess HCl
NaHCO3 (s) + H+ (aq)  Na+ (aq) + H2O (l) + CO2 (g)
+ Cl- (aq)
+ Cl- (aq)
NaCl(s)
 Na+ (aq) + Cl-(aq)
(NaCl will dissolve in, but not react with, HCl)
3. Capture Liberated Gas and Measure its volume, vCO2
7
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
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But - it Involves Some Important Concepts
Limiting Reagents
Gas Mixtures – Partial Pressure
Gas Solubility – Henry’s Law
Does CO2 behave as an ideal gas?
si = kHPi
Accuracy & Reproducibility
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.
A. smaller than
B. equal to
C. larger than
0%
A.
0%
B.
0%
C.
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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
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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. Assume unknown is pure NaHCO3
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
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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
NaHCO3 (s) + H+ (aq) Na+ (aq) + H2O (l) + CO2 (g)
analytical
balance
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)
At STP: 1 mole occupies 22.4 L
1 mmole occupies 22.4 mL
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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,
(200 mg NaHCO3 = 2.4 mmol)
We produce at most 2.4 mmol of liquid H2O
Is this volume significant compared to the
~ 60 mL of CO2 produced?
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The volume of 2.4 mmol of liquid H2O is
A. about 1 mL
B. about 1 drop
C. a few mL
0%
A.
0%
B.
0%
C.
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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
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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
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When reaction is complete, all the sodium in the
initial sample is present as aqueous NaCl.
A. True
B. False
0%
A.
0%
B.
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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)
ClCl-
The samples consist of NaHCO3 and NaCl.
In the reaction with HCl, all of the
NaHCO3 is converted to NaCl
A = True
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The Apparatus
EXTENSION
CLAMP
RIGHT
ANGLE
ELBOW
SYRINGE
LARGE
TEST
TUBE
SMALL
TEST
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TUBE
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
0 mmol
Air (g)
w mmol
w mmol
NaCl (aq)
0 mmol
x + y mmol
CO2 (g)
0 mmol
x - z mmol
CO2 (aq)
0 mmol
10 mL of 1.0 M HCl contains:
10 X 1.0 = 10 mmol HCl
z mmol
NaHCO3 (s) + H+ (aq) + Cl- (aq)  Na+ (aq) + H2O (l) + CO2 (g) + Cl- (aq)
10Some
mL H2O of
= 10the
g H2OCO will
2
10 g = 10,000 mg / 18.0 mg/mmol
dissolve
in Hthe
water
= 556 mmol
2O
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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
And after reaction,
• the liberated CO2
and
Patm
And a small amount of
non-volatile stuff:
200 mg of NaHCO3/NaCl
PH2O + Pair = Patm
PH2O, Pair
PCO2
PH2O + Pair + PCO2 = Patm
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What is the Magnitude of PH2O?
Table showing vapor pressure of water as a
Vapor Pressure of Water
function of temperature is posted in lab.
PH2O is a
function of
temperature
Vapor Pressure (mm Hg)
60
50
40
30
20
10
0
0
5
10
15
20
25
30
35
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Temperature (oC)
Over the range 20oC – 30oC, PH2O increases from:
Pressureto
is 31.8
often
in mm of Hg (Torr)
mmmeasured
Hg
17.5
(0.023 to 0.042 atm)
1 atm = 760 mm Hg
2.3% to 4.2% for P ~1 atm
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P
PPiH2O
H2O ,T, vi, nair
f
PPfH2O
,T,
v
,
n
,
P
H2O
f
air
CO2
P
nCO2g
nCO2s H2O(l)
H2O(l)
Initially, we have
P = Piair + PiH2O
Finally, we have
P = Pfair + PfH2O + PfCO2
P,T & nair don’t change, so
PiH2O = PfH2O = PH2O and
P – PH2O = Piair
P = n RT/v
P – PH2O = Pfair + PfCO2
Piair = Pfair + PfCO2
PfCO2 = Piair - Pfair
nCO2g RT/vftot = nair RT/vitot - nair RT/vftot
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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?
PFCO2 = nCO2g RT/vftot
PfCO2 = [(P - PH2O) (vftot - vitot) /RT ] RT/vftot
PfCO2 = (P - PH2O) (vftot – vitot)/vftot
Equation 6
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What are the various volumes in the exercise?
vtube: The volume of
just the large test
tube and the right
angle elbow
vitot & vftot: The gas
phase* volume of the
entire closed system
before & after the
reaction
vtube
You will
measure
this!
vtube – vHCl + visyringe
*We must exclude the volume of the liquid HCl but can ignore the solid NaHCO3
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NOTEthat:
THAT:
Note
Equation 4
vftot – vitot
= vtube – vHCl + vfsyringe – vtube + vHCl – visyringe
vF – vI = vSYS – vHCl + vfsyringe – vSYS + vHCl – visyringe
BUT
Equation 6
1 - vitot / vftot
The most common
errir in this exercise is
to use syringe volumes
in equation 6
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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.
kH Given in
SUSB-054
SCO2 = kH * PCO2 ( kH = 3.2 X 10-2 mol / L-atm )
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
From the
table of
PH2O vs T
• P = 1.000 atm,
• T=
22oC
PH2O =
20 mmatm
0.026
Hg
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PCO2 =
( P - PH2O ) ( 1 – vitot / vftot)
vitot =
95.0 mL – 10.0 mL + 5.0 mL
Equation 6
=
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
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What percentage error does 0.1 mmol represent
out of the typical 3.0 mmol of CO2 liberated in
this exercise
A. 0.33 %
B. 3.3 %
C. 33. %
0%
A.
0%
B.
0%
C.
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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%
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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 +v a / v2 ) ( v – b ) = RT
1 mol
For CO2 at Room Temperature, the corrections
to P and v are:
a / v2 b / v
CO2
0.6%
0.18%
Small corrections compared to others
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Review
Affected
Variable
Effect
%
Volume of H2O Produced
vCO2g
0.1
X
Non -Ideality of CO2
PCO2
0.6
X
Vapor Pressure of Water
PCO2
Solubility of CO2
nCO2
Issue
2.3 – 4.2 
3.0
If we seek accuracy to within 1%
in the % NaHCO3

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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%
(0.5 / 50)
~1%
vCO2g
syringe
Precision should be about 1%
Accuracy should be less than 3%
* Assuming no ambient temperature changes – see Prob 2
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Calculations
vCalculated
Measured
Posted
- vinit
final
Weight of Sample
Volume of gaseous CO2
v
Pressure,
P = 752 mm Hg =
Temperature, T = 23oC =
[email protected] 23oC (from Table) 21 mm Hg
0.2147 g
45.7 mL
0.989 atm
296 K
0.028 atm
o=
mmol CO2 (gas) Initial
= (P
-mm
PCH2O
)v =/mL
RT
752
23
Hg
==5.0
21
mmHg
Final
50.7
mmol CO2 (liquid)
(Henry’s
Law)
23
752+ =/273
760
K mL
21/760
= 296
atmatm
K
= 0.028
Tot CO2
= 0.989 atm
1.81 mmol
0.10 mmol
1.91 mmol
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To calculate the Henry’s Law correction, we need the
total volume of the system. Suppose vtube is 100.0 mL
That makes the initial volume,
vitot = 100.0 - 10.0 + 5.0
= 95.0 mL
and the final volume,
vftot = 100.0 - 10.0 + 50.7
= 140.7 mL
vtube = 100.0 mL
Syringe:
Initial = 5.0 mL
Final = 50.7 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
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Calculations
Calculated
Weight of Sample
Volume of gaseous CO2
v
Pressure,
P = 752 mm Hg =
Temperature, T = 23oC =
[email protected] 23oC (from Table) 21 mm Hg
0.2147 g
45.7 mL
0.989 atm
296 K
0.028 atm
mmol CO2 (gas) = (P - PH2O)v / RT
mmol CO2 (liquid) (Henry’s Law)
1.81 mmol
0.10 mmol
Tot CO2
1.91 mmol
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 %38
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.
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CHE 133
MAKE-UP LABORATORY EXERCISE
Mon 11/19 - 2:00 PM or Tue 11/20 – 2:30 PM
Eligible Students are ONLY those who have
excused absences from
either a TEST (105 point)
or
PRELIMINARY (55 point) Exercise
IN EACH LABORATORY ROOM
BY November 15
Everyone does the same make-up exercise. You can download the
exercise at:
http://www.ic.sunysb.edu/Class/che133/susb/SUSB055.pdf 40
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
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``` # ENVI 1106 SAMPLE MIDTERM 3 QUESTIONS Questions 1-4 (circle the correct answer): # SAMPLE DIABETIC KETOACIDOSIS (DKA) CLINICAL ORDER SET Adult 