Determination of the Percent Acetic Acid in Vinegar

Experiment 4

SOLUTION STOICHIOMETRY

ACID-BASE TITRATIONS

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) H2SO4 + 2NaOH ------> Na2SO4 + 2HOH

(3) 2HCl + Ca(OH)2 ------> CaCl2 + 2HOH

(4) H2SO4 + Ca(OH)2 ------> CaSO4 + 2HOH

(5) 2H3PO4 + 3Ca(OH)2 ------> Ca3(PO4)2 + 6H2O

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 acid-base 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 H2SO4 (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 H2SO4 reacts with 2 moles of NaOH (or one molecule of H2SO4 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

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 (H2SO4). In this volume there are (by eq 7) na 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) na = (0.13 mmol H2SO4/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, Vb, of the NaOH required to react with the 34.56 mL of 0.13M H2SO4 solution is given by

(9) nb = (0.21 mmol/mL)(Vb ) = 2(4.493) mmol

(0.21 mmol/mL)Vb = 2(4.493)mmol

and (10) Vb = 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 H2SO4 solution. If these volumes are mixed, one gets a solution of Na2SO4 in water.

What is done stepwise in equations (8) through (10) for reaction (2) can be written as a single equation

(11) 2MaVa = MbVb

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 H2SO4 (see equation (2) above).

Substitution of the values in eq 11 gives

(11A) 2(0.13mmol/mL)(34.56mL) = (0.21 mmol/mL)Vb

and the value for Vb is the same as in eq 10.

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: MaVa = MbVb

(13) reaction 3: MaVa = 2MbVb

(14) reaction 4: MaVa = MbVb

(15) reaction 5: 3MaVa = 2MbVb

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.

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) NaVa = NbVb

For example, there are 2 hydroxides in a molecule of Ca(OH)2 and there are 3 hydrogens in a molecule of H3PO4. Therefore

(18) for Ca(OH)2 Nb = 2Mb

(19) for H3PO4 Na = 3Ma

Using (18) and (19) in (15) gives (17). Proceeding similarly for the other reactions will also give (17).

From equation (16)

and rearrangement gives

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) H2SO4 (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 H2SO4 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 Vi milliliters of 6M solution is 6Vi. The number of mmoles of NaOH in Vf milliliters of 0.2M solution is 0.2Vf. 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) 6Vi = 0.2Vf

One must choose the amount of 0.2M solution desired. Suppose that 300 mL are needed. Then

(27) 6Vi = (0.2)(300)

so that Vi = 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 MiVi = MfVf for 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 H2C204

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, H2C2O4, 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) H2C2O4 + 2NaOH ------> Na2C2O4 + 2HOH

therefore Na = 2Ma and Nb = Mb and one can solve the problem using two methods:

USING MOLARITY USING NORMALITY

(29) 2MaVa = MbVb NaVa = NbVb

so that

(30) 2(0.1)(34.56) = Mb(28.67) (0.2)(34.56) = Nb(28.67)

and

(31) Mb = 0.24 moles/liter Nb = 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 (HC2H3O2 in water) requires 77.10 mL of the NaOH solution standardized above to neutralize it. What is the weight percent acetic acid (HC2H3O2) in the vinegar?