34
Experiment 4
SOLUTION STOICHIOMETRY
ACID-BASE TITRATIONS
Determination of the Percent Acetic Acid in Vinegar
Objectives: (1) To introduce and use the concept of solution stoichiometry
(2) To specifically use solution stoichiometry to determine the percent of acetic acid in
vinegar.
(3) To prepare a standard solution by the method of titration.
Consider the following balanced chemical equations:
(1) HCl + NaOH ------> NaCl + HOH
(2) H
2
SO
4
+ 2NaOH --------> Na
2
SO
4
+ 2HOH
(3) 2HCl + Ca(OH)
2
---------> CaCl
2
+ 2HOH
(4) H
2
SO
4
+ Ca(OH)
2
---------> CaSO
4
+ 2HOH
(5) 2H
3
PO
4
+ 3Ca(OH)
2
----------> Ca
3
(PO
4
)
2
+ 6H
2
O
These reactions in water have one feature in common: a hydrogen ion from one compound reacts with a hydroxide ion from the other compound to form water. The compound furnishing the hydrogen ion is called an acid and the one furnishing the hydroxide ion is called a base. The metal ion (the cation) of the base and the anion of the acid combine to form a salt. In this context, salt is essentially synonymous with ionic compound. Reactions such as these are called acidbase reactions.
The concepts of solution reactions and stoichiometry in this experiment are applicable to any chemical reaction taking place in solution, not just the acid base reactions indicated here.
Stoichiometry is the area of chemistry that deals with how much of one compound reacts with another. When compounds are mixed in amounts such that these amounts just exactly react and none of any reactant is in excess (they are all the limiting reagent), this mixture is said to be a stoichiometric mixture or that stoichiometric amounts have been mixed. Observation of the above balanced equations indicates that stoichiometrically reaction occurs so that the total number of hydrogen ions available in the amount of acid reacting is the same as the total number of hydroxide ions available in the amount of base reacting. Thus in reaction (2) above, one mole (or one molecule) of H
2
SO
4
(sulfuric acid) can furnish the same number of hydrogens as the number of hydroxides that two mole (or two molecules) of NaOH can furnish. Therefore one mole of
H
2
SO
4
reacts with 2 moles of NaOH (or one molecule of H
2
SO
4
reacts with two molecules of
NaOH). This type of stoichiometry information is obtained from any balanced chemical equation.
In a solution the molarity, M, is the number of moles of solute in a liter of solution. Thus
(6) M = n/V

35 where n is the number of moles and V the volume in liters. By equation (6), the number of moles of solute in V liters of a solution of molarity M is
(7) n = MV
Suppose one takes 34.56 mL of a 0.13 M solution of sulfuric acid (H
2
SO
4
). In this volume there are (by eq 7) n a
moles of sulfuric acid
The units for molarity are normally moles/liter, however, sometimes (especially if a person is using a very low concentration), units of molarity can be millimoles/milliliter or simply mmol/mL. The numerical value of the concentration is the same, just the units are different. i.e. convert 0.13 moles/liter to mmol/mL
(0.13 mol/mL)(1000 mmol/1 mol)(1L/1000mL) = 0.13 mmol/mL
*note that in the conversion, both the numerator and denominator are multiplied by 1000; thereby cancelling each other out.
(8) n a
= (0.13 mmol H
2
SO
4
/mL)(34.56 mL) = 4.493 mmol
SEE EXPERIMENT 2 FOR THE ABOVE DISCUSSION IN THE CONTEXT OF SOLUTION
PREPARATION.
In order to react with 4.493 mmoles of sulfuric acid, one needs 2(4.493) millimoles of NaOH according to reaction (2) above. If one has a 0.21 M solution of NaOH then the volume, V b
, of the NaOH required to react with the 34.56 mL of 0.13M H
2
SO
4
solution is given by
(9) n b
= (0.21 mmol/mL)(V b
) = 2(4.493) mmol
(0.21 mmol/mL)V b
= 2(4.493)mmol
and (10) V b
= 42.79 mL
Therefore, if reaction occurs as in (2) above, it will require 42.79 mL of 0.21 M NaOH solution to neutralize (stoichiometrically react with) 34.56 mL of 0.13 M H
2
SO
4
solution. If these volumes are mixed, one gets a solution of Na
2
SO
4
in water.
What is done stepwise in equations (8) through (10) for reaction (2) can be written as a single equation
(11) 2M a
V a
= M b
V b where the subscript “a” indicates acid and the subscript “b” indicates base. The factor 2 in equation (11) is a result of the fact that the number of moles (or millimoles) of NaOH required is

two (2) times the number of moles (or millimoles) of H
2
SO
4
(see equation (2) above).
Substitution of the values in eq 11 gives
(11A) 2(0.13mmol/mL)(34.56mL) = (0.21 mmol/mL)V b and the value for V b
is the same as in eq 10.
36
If one has a certain volume of sulfuric acid solution, there is one volume of a given NaOH solution needed such that equivalent amounts of the two reactants are mixed. Equivalent amounts are the amounts that stoichiometrically react (that is, none of either reactant is left over). When the amount of NaOH (in solution or otherwise) needed for stoichiometric reaction has been added, this is called the equivalence point.
In the laboratory, the determination of the volume (the amount) of NaOH required to react stoichiometrically with the sulfuric acid solution is done by a procedure called titration. A burette is used to measure an accurate volume of the sulfuric acid solution into a flask. A few drops of an indicator is added and another burette is used to add the NaOH solution to the sulfuric acid solution until there is a visual change in color of the solution (due to the indicator).
There is a visual change in color in the solution caused by the indicator, a substance that changes color as close as possible to the point when the stoichiometric amount of NaOH (in this case) has been added. When the solution changes color (the indicator changes color) this is called the endpoint of the titration. The solution should be colorless and change to a pink color
(for the phenolphthalein indicator used here) upon the addition of one drop of NaOH titrant .
The indicator phenolphthalein is pink in basic solutions and colorless in acid solutions. The solution for the titration of sulfuric acid with NaOH changes from acidic (indicator colorless) to basic (indicator pink) at the endpoint. At the endpoint the solution is slightly basic when the titrant is a base. Other indicators have a different color change but work the same in principle.
The endpoint and the equivalence point should be as nearly the same as possible. The correct indicator must be chosen to make that be so. That choice has been made for this experiment and is phenolphthalein.
Eq 11 (and eqs 12 to 15 below) are valid only at the equivalence point since that is the only place in the titration that the number of moles of each reactant are present in the amounts required for reaction by the balanced equation (that is, a stoichiometric mixture).
Eqs 12 TO 15 BELOW ARE SIMILAR IN FORM AND APPEARANCE TO THE EQUATIONS
USED FOR DILUTION (ESPECIALLY Eqs 12 AND 14) BUT THE CONCEPTUAL
MEANINGS ARE DIFFERENT.
Equation (11) is for reaction (2) above. The form of (11) changes slightly for different acid base reactions. The equations equivalent to equation (11) for reactions (1), (3), (4) and (5) above are:
(EQN 12 TO 15 ARE VALID ONLY AT THE EQUIVALENCE POINT)

(12) reaction 1: M a
V a
= M b
V b
(13) reaction 3: M a
V a
= 2M b
V b
(14) reaction 4: M a
V a
= M b
V b
(15) reaction 5: 3M a
V a
= 2M b
V b
In each equation, note that the product of molarity and volume of the acid is multiplied by the coefficient of the base from the balanced equation and that the product of the molarity and volume of the base is multiplied by the coefficient of the acid from the balanced equation.
37
If one defines a new concentration unit, the normality (symbol N), as
(16) N = zM, where z is the number of acid hydrogens in a molecule of an acid (for acid solutions) or the number of hydroxide ions in a molecule of a base (for base solutions), then equations (11) through (15) can all be written as one equation:
(17) N a
V a
= N b
V b
For example, there are 2 hydroxides in a molecule of Ca(OH)
2
and there are 3 hydrogens in a molecule of H
3
PO
4
. Therefore
(18) for Ca(OH)
2
N b
= 2M b
(19) for H
3
PO
4
N a
= 3M a
Using (18) and (19) in (15) gives (17). Proceeding similarly for the other reactions will also give
(17).
From equation (16) gms solute
(20) N = zM = z molar mass
V and rearrangement gives

38
(21) N = gms solute molar mass
V z
= gms solute
GEW
V
In equation (21) the molar mass divided by z is the weight of the acid or base that contains a mole of hydrogen (for an acid) or a mole hydroxide (for a base). This follows from the definition of z.
THE WEIGHT OF AN ACID IN GRAMS THAT WILL FURNISH ONE MOLE OF HYDROGEN
IONS IS THE GRAM EQUIVALENT WEIGHT (GEW) OF THE ACID. THE WEIGHT OF A
BASE IN GRAMS THAT WILL FURNISH ONE MOLE OF HYDROXIDE IONS IS THE GRAM
EQUIVALENT WEIGHT (GEW) OF THE BASE.
Examples of calculation of equivalent weights are
(22) HCl (z = 1) (molar mass)/z = 36.5/1 = 36.5 grams - contains one mole of H +
(23) H
2
SO
4
(z = 2) (molar mass)/z = 98/2 = 49 grams -contains one mole of H +
(24) Ca(OH)
2
(z = 2) (molar mass)/z = 74/2 = 37 grams - contains one mole of OH -
(25) NaOH (z = 1) (molar mass)/z = 40/1 = 40 grams - contains one mole of OH -
Thus 36.5 grams of HCl will contain 1 mole of hydrogen; 49 grams of H
2
SO
4
will contain 1 mole of hydrogen. 37 grams of Ca(OH)
2
and 40 grams of NaOH will each contain 1 mole of hydroxide. The weight of any acid that contains 1 mole of hydrogen will react with the weight of any base that contains 1 mole of hydroxide. These weights are called equivalent weights because they just exactly react with one another, that is, they are equivalent in reaction.
Q EQUIVALENT WEIGHTS OF ANY ACID REACTS WITH Q EQUIVALENT WEIGHTS OF
ANY BASE FOR ANY NUMBER Q, THAT IS, REACTION IS ON A 1 TO 1 BASIS IF
EQUIVALENTS ARE USED. THIS IS NOT TRUE FOR MOLES.
IF MOLES AND MOLARITY ARE USED, A BALANCED EQUATION IS NECESSARY TO
DETERMINE THE RATIO IN WHICH REACTION OCCURS.
DILUTIONS (SEE EXPERIMENT 2 FOR THIS DISCUSSION, ALSO)
Suppose one wishes to prepare a solution of NaOH that is 0.2 M from a stock solution that is
6M in NaOH. From equation (7) the number of mmoles of NaOH in V i
milliliters of 6M solution is
6V i
. The number of mmoles of NaOH in V f
milliliters of 0.2M solution is 0.2V
f
. If only water is used to dilute the 6M solution, then the number of mmoles of NaOH in the final diluted solution is the same as the number of mmoles of NaOH in initial volume of 6M solution used. Thus
(26) 6V i
= 0.2V
f
One must choose the amount of 0.2M solution desired. Suppose that 300 mL are needed.
Then

(27) 6V i
= (0.2)(300) so that V i
= 10 mL. Thus one dilutes 10 mL of 6M NaOH to a total final volume of 300 mL with water to obtain a 0.2 M solution. The general equation used here is M i
V i
= M f
V f
for
39 dilution with solvent.
The 0.2 M solution obtained here is approximate unless the concentration of the solution diluted is known very accurately and the volumes used are measured accurately using volumetric flasks, etc. These conditions were met in Experiment 2 where standard solutions were made by the direct method and by dilution. In the case here, the concentration of the 6M solution is known only approximately and the volumes in dilution are measured approximately so the concentration of 0.2 M for the diluted solution is approximate. The accurate value will be determined by titration with a standard solution. A standard solution is one whose concentration is accurately known.
The 6M NaOH solution cannot be made accurately by the direct method used in Experiment 2 because NaOH is hygroscopic (it absorbs water from the air) and an accurate sample cannot be weighed. Standard NaOH solutions are usually prepared by the titration method used in this experiment.
EXAMPLE CALCULATION: DETERMINATION OF THE ACETIC ACID IN VINEGAR
I. STANDARDIZATION OF THE NaOH SOLUTION USING STANDARD H
2
C
2
0
4
A standard solution is one whose concentration is known very accurately. The process of determining the accurate concentration is called standardization.
Suppose that one has a 0.100 M solution of oxalic acid, H
2
C
2
O
4
, and 34.56 mL of this solution will neutralize 28.67 mL of the NaOH solution prepared by dilution above. One can use this data to determine the normality and/or the molarity of the NaOH solution as follows:
For the reaction of the oxalic acid and NaOH,
(28) H
2
C
2
O
4
+ 2NaOH --------> Na
2
C
2
O
4
+ 2HOH therefore N a
= 2M a
and N b
= M b
and one can solve the problem using two methods:
USING MOLARITY USING NORMALITY
(29) 2M a
V a
= M b
V b
N a
V a
= N b
V b so that
(30) 2(0.1)(34.56) = M b
(28.67) (0.2)(34.56) = N b
(28.67) and
(31) M b
= 0.24 moles/liter N b
= 0.24 eq/lilter
= 0.24 mmol/mL = 0.24 meq/mL
Since the NaOH has one hydroxide, the molarity and the normality are the same.
II. TITRATION OF THE VINEGAR WITH THE NaOH

It is found that a 21.58 gram sample of vinegar (HC
2
H
3
O
2
in water) requires 77.10 mL of the
NaOH solution standardized above to neutralize it. What is the weight percent acetic acid
40
(HC
2
H
3
O
2
) in the vinegar?
For the reaction of acetic acid and NaOH:
(32) HC
2
H
3
O
2
+ NaOH ---> NaC
2
H
3
O
2
+ HOH
I. BECAUSE NaOH HAS ONE OH PER FORMULA UNIT:
(32A) # of mmoles of NaOH used in titration = # of milliequivalents of NaOH used in titration
II. BECAUSE THE NUMBER OF MILLIEQUIVALENTS OF BASE MUST EQUAL THE NUMBER OF
MILLIEQUIVALENTS OF ACID AT THE EQUIVALENCE POINT (FOR ANY ACID REACTING WITH
ANY BASE):
(32B) # of mmol of NaOH used in titration = # of milliequivalents of HC
2
H
3
O
2
in vinegar sample
II. BECAUSE HC
2
H
3
O
2
HAS ONE H + PER MOLECULE:
(32C) # of mmol of NaOH used in titration = # of mmol of HC
2
H
3
O
2
in the sample
Eq 32C is seen to be true from the balanced equation (eq 32) also.
(33) # of mmol of NaOH used = M b
V b
= (0.24)(77.10) = 18.6 mmol NaOH
Therefore by eq 32C
(33A) # mmol of acetic acid in sample = 18.6 mmol
(34) grams of HC
2
H
3
O
2
= (# of moles of HC
2
H
3
O
2
)(molar mass of HC
2
H
3
O
2
)
= (0.0186)(60) = 1.11 grams of acetic acid in sample
Eqn 34 can be algebraically manipulated
(34A) = [(1000)(# of moles of HC
2
H
3
O
2
)](molar mass of HC
2
H
3
O
2
)
(1000)
(34B) = (# of mmoles of HC
2
H
3
O
2
)(mmolar mass of HC
2
H
3
O
2
)
The millimolar mass of any compound is the mass of 1 millimole = MM/1000 = 60/1000 for acetic acid, so
(35) grams of acetic acid = (18.6)(0.060) grams of acetic acid= 1.11 grams of acetic acid
The percent acetic acid by weight in the sample is (by definition of percent)
(36) % acetic acid by weight = (gms acetic acid) (100)
gms of vinegar sample

41
(37) = (1.11) (100) = 5.2%
21.58
The general equation for % acetic acid is (volumes in milliliters)
(38) % acetic acid = (N b
)(V b
)(mGEW of acetic acid)(100)
(sample wt.)
(39) = (M b
)(V b
)(mMM of acetic acid)(100)
(sample wt) where eq 39 follows from eq 38 because acetic acid has one acid hydrogen and reacts one to one with
NaOH which has one hydroxide. Therefore the molarity of the base is equal to the normality of the base and the mGEW of acetic acid is the same as the millimolar mass of acetic acid.
In general for a sample of an acid which has z acid hydrogens titrated with NaOH, the equation using normality, eq 38, is unchanged but eq 39 becomes eq 40
(40) % acid = (1/z)(M b
)(V b
)(mMM of acid)(100)
(sample wt) mGEW of acid
% acid = (MaVa)(mMM of acid/z)(100) sample wt
For a sample of an acid with z acid
acid hydrogens titrated with a base with z base
hydroxide ions, the percent of the acid in the sample is given by
(41) % acid = (z base
/z acid
)(M b
)(V b
)(mMM of acid)(100)
(sample wt.)
N
NaOH mGEW of acid
% acid = (z base
MbVb)mMM of acid/z acid
)(100) sample wt

42
Titrations are usually used to determine the amount of a substance in a sample of a solid or in a solution by determining the volume of titrant that is needed to react with the desired compound. In order to do this, the molarity and/or normality of the titrant solution must be accurately known and the chemical reaction between them must be known.
This concepts and calculations above are not restricted to acid/base reactions and can be used for other types of chemical reactions. The definition of gram equivalent weight may change, however, to keep the requirement that one equivalent react with one equivalent. See the appendix to Experiment 9 for more illustration of this.
If one considers the operations above, it will become clear why solutions are used in the laboratory. It would be difficult without using solutions to effect the reaction of the sodium hydroxide (a solid) with the vinegar sample (a liquid) and make accurate measurements in regard to how much sodium hydroxide would be required to react with a given sample of the vinegar. Using solutions makes the process much easier and convenient and is one of the reasons chemists use solutions. It does, however, require that one learn about solution concentrations and how they can relate to reaction stoichiometry.
ACID HYDROGENS
In eq 32, note that in reaction only one of the four hydrogens in acetic acid reacts with the NaOH to give a salt and water. Not all the hydrogens in compounds will react with the hydroxide of NaOH to form water.
Those that will are called acid hydrogens and the acid hydrogens are written first in the formula for a compound. Thus the formula for acetic acid is HC
2
H
3
O
2
indicating that it has one acid hydrogen. The determination of which and how many of the hydrogens are acid hydrogens in a compound must be done experimentally but once this is done, the formula is written so as to indicate this.
CRITERION FOR REJECTING RESULTS
Suppose one determines the normality of a NaOH solution 5 times. The results are shown in the second column of Table I. In the third column of Table I is the difference in the normality of each individual run and the arithmetical average value of the normality.
This difference can be viewed as the deviation o r “error” of each run from the average. In the fourth column is the parts per thousand relative error (ppth) for each run. Parts per thousand relative error is calculated from the equation
parts per thousand relative error = (N n
- N ave
)(1000)
N ave

Table 1.
Run # Normality
1
2
3
4
5
0.2058
0.2065
0.2096
0.2050
0.2010
N ave
= 0.2056
N n
- N ave
+0.0002
+0.0009
0.0040
-0.0006
-0.0046 parts per thousand relative error
1
4
19
3
22
43
It is found that in the laboratory one can routinely obtain results in this type of procedure such that the average and individual runs differ in the order of 2 to 3 parts per thousand. Therefore the deviations from the average of 19 and 22 ppth are large enough to justify rejecting these values. One would average the other three values S(runs 1,2 and 4)and use the average as the accepted value of the normality.
PROCEDURE:
I. Preparation and Standardization of the NaOH solution
1. (a) In a 500 or 600 mL Erlenmeyer flask, prepare 350 mL of a NaOH solution to be approximately 0.2 M by diluting a concentrated solution (6M) of NaOH. Use the relationship
M i
V i
= M f
V f
as done in eqs 26 and 27 above. The volume of NaOH can be measured using a graduated cylinder here. If a glass cylinder is used, rinse it well with water after use.
SWIRL THE NaOH SOLUTION WELL TO MIX BEFORE USING.
(b) Rinse and fill a buret with the 0.2M solution of NaOH. See the procedure in Experiment
1.
2. Obtain about 100 mL of a standard oxalic acid solution (a standard solution is one of accurately known molarity) from the container in the laboratory. Rinse and fill another buret with this solution of oxalic acid.
3. Take and record an initial reading on both burets. READ THE BURET TO 0.01 mL (TWO
DECIMAL PLACES) AND RECORD TWO DECIMAL PLACES. From the buret filled in step 2, run about 20 to 25 mL of the oxalic acid solution into a 125 (a 250 mL flask is OK) mL
Erlenmeyer flask. Add about 10 to 15 mL distilled water and a few drops of phenolphthalein indicator solution.
4. Titrate the oxalic acid with the NaOH solution (while swirling the solution in the flask) until the point is reached at which one drop of the NaOH solution changes the color of the solution from

44 colorless to pink. Just before the color changes, wash down the sides of the flask with distilled water used a wash bottle to be sure all reactants are in the solution.
This is the endpoint, the point at which the indicator changes color. The indicator is chosen so that it changes color when essentially equivalent amounts of the acid and base have been added. Neutralization has occurred at this point and one has a stoichiometric mixture.
If too much NaOH is added, add some more (that is, back titrate with the) oxalic acid solution and then titrate with the NaOH until the endpoint is again reached. Take final readings of both burets only when a satisfactory endpoint has been achieved.
5. Pour the solution from step 4 in the waste container, rinse the flask well, and repeat steps 3 and 4 at least twice. Refill the burets when necessary. For each run, calculate the molarity of the NaOH solution from the known value of the molarity or normality of the acid using eq 29.
Average the values for all the good runs (at least two) of the molarity. These values should not differ by more than 5 to 10 parts per thousand relative from the average (which amounts usually to 1 to 2 units in the thousands place).
II. Analysis of Vinegar for % Acetic Acid
6. Rinse a 250 mL beaker and dry the outside . Place the beaker on the balance, tare and add vinegar to the beaker (with a dropper is slow, but sure) to get approximately accurately 7 grams of vinegar sample. Add 15 to 20 mL of distilled water to the flask and 2 to 3 drops of phenolphthalein indicator solution.
7. Place the beaker on a magnetic stirrer on a ring stand as in Figure 1 and add a stirring bar. If no magnetic stirrer is available, you need to stir with a stirring rod during the titration.
FIGURE 1
8. Refill the NaOH burette from the standardization above to slightly above a 0.00 reading.
Drain NaOH solution into a waste beaker very slowly until the reading is 0.00. Discard the
NaOH in the beaker in the waste containers. Use a burette clamp to attach the burette to the ring stand as indicated in Figure 1 above
Record the precise concentration of the NaOH solution in your data table.
9. Use a utility clamp to suspend a pH Sensor on a ring stand as shown in Figure 1. Position the pH Sensor in the vinegar solution and adjust its position toward the outside of the beaker so it will not be struck by the stirring bar. Turn on the magnetic stirrer, and adjust it to a medium stirring rate (with no splashing of solution).
10. Connect the pH Sensor to the computer interface. Prepare the computer for data collection by opening the file “Experiment #7 Diprotic Acid Titration” from Logger Pro .

45
11. You are now ready to begin the titration. This process goes faster if one person manipulates and reads the burette while another person operates the computer and enters burette readings. a Before adding NaOH titrant, click has stabilized, click and monitor pH for 5-10 seconds. Once the pH
.
In the edit box, type “0” (for 0 mL or drops added), and press
ENTER to store the first data pair for this experiment. There should appear on the graph a dot to represent this data pair. There is a “current” dot coincident with this one but they appear as only one dot. b Add one drop of NaOH. If there is little or no movement of the red dot, add NaOH slowly
(rapid drop-wise) and watch the dot. The dot for the first data pair remains stationary but the current dot moves to higher pH slowly. When the dot moves so that it is just above the stationary dot (may take several mL), stop adding NaOH. When the pH stabilizes, again click . In the edit box, type the current burette reading, to the nearest 0.01 mL. Press
ENTER. You have now saved the second data pair for the experiment. A second dot will appear on the graph for this data pair. Coincident with this dot is a dot that will later become the dot for the third data pair. c Repeat step b. For the first few dots several milliliters will be needed. For 7 grams of vinegar and 0.2M NaOH, this will change in the neighborhood of 30 mL. For volumes above about 25 mL pay attention to what happens after the first drop of NaOH. At some volume this one drop will cause a relatively large change in pH. Here one or maybe two drops is all that is needed between saving data points. This will happen for perhaps a couple of data points d After the rapid rise region the graph will curve back toward the horizontal and several milliliters of NaOH will again be needed to move the dot as in part b. Get two or three data points after it begins to curve toward horizontal. e When data collection is complete, click . Dispose of the beaker contents as directed by the instructor. PRINT A COPY OF THE DATA TABLE AND THE GRAPH.
Determine the volume of NaOH AT THE EQUIVALENCE POINT AND CALCULATE THE %
VINEGAR.
Note: It is desired to find the exact location on the graph where the change in pH is a maximum. This is indicated when the second derivative is closest to zero. This is displayed in Logger Pro as follows. a. Select connect points on graph. b. Select display second derivative c. Drop down menu – Analyze – select Examine and Interpolate. d. Move cursor to point at which second derivative is closest to zero and record value in box.

DATA SHEET
I. Determinaltion of the NaOH molarity Run 1 Run 2 Run 3
Molarity of Oxalic Acid (look on container in lab) ________
Final burette reading (oxalic acid) _________ _________ _________
Initial burette reading (oxalic acid) _________ _________ _________
Volume (mL) of oxalic acid used _________ _________ __________
(V a
in eq. 29)
Final burette reading (NaOH) _________ _________ __________
Initial burette reading (NaOH) _________ _________ __________
Volume (mL) of NaOH used __________ _________ __________
(V b
in eq 29)
Molarity of NaOH solution (from eq 29) __________ _________ __________
Average molarity of NaOH _________
II. Determination of Percent acetic acid in vinegar
Mass of vinegar sample _________ _________ ___________
Final burette reading of NaOH _________ _________ _________
Initial burette reading of NaOH _________ _________ __________
Volume (mL) of NaOH used _________ _________ __________
(V b
in eq 33) millimoles of NaOH used in titration _________ _________ __________
(eq 33) millimoles of acetic acid in vinegar sample _________ _________ ___________ grams of acetic acid in vinegar sample _________ _________ __________
(eq 34)
Percent acetic acid in vinegar sample _________ _________ ________
Average percent acetic acid in sample __________
46