1. LAB 1

DETERMINATION OF THE CONCENTRATION OF ACETIC ACID IN VINEGAR

1.1 Introduction

Concentration of solution is the amount of solute in a given amount os solvent. A concentrated

solution contains relatively large quantity of solute in a given amount of solvent. Dilute solutions

contain relatively little solute in a given amount of solvent. There are 2 specifics term to express

concenrtation, which is molarity and percent by mass.

Molarity is the number of moles of solute per liter of solution.

( )moles of solute

Molarity Mliter of solution

(Equation 1-1)

Percent by mass is the mass in grmas of solute per 100 grams of solution

100%grams of solute

Percent solute xgrams of solution

(Equation 1-2)

Vinegar is a dilute solution of acitic acid. The molecular formula for acetic acid is CH3COOH.

Both molarity and percent by mass of aceric acid in a vinegar solution can be determine by

performing a titration. A titration is a process in which small increments of a solution of known

concerntration are added to a specific volume of a solution of unknown concerntration until the

stoichiometry for that reaction is attained. Knowing the quantity of the known solution required

to complete the titration, calculation of the unknown solution ca be done. The purpose of titration

is to determine the equivalance point of the reaction. The equivalance point is reach when the

added quantity of one reactant is the exact amount necessary for stoichiometric reaction with

another reactant.

1.2 Objectives

To :

(a) Determine the morality of a solution and the percent by mass of acetic acid in vinegar by

titration with the standardized sodium hydroxide solution.

1.3 Theory

In the titration process, a burette is used to dispense a small, quantifiable increment of solution of

known concentration (Figure 1.1). A typical burette has the smallest calibration unit of 0.1mL

(Figure 1.2), therefore, volume dispense from the burette should be estimated to the nearest

0.01mL.

Figure 1-1: a) Depicts a typical 50-mL burette. b) Indicates smallest calibration unit, 0.1mL, on a typical 50 mL burette

In this experiment, the equivalence point occurs when the moles of acid in the solution equals to

the moles of base added in the titration. For example, the stoichiometric amount of 1 mole of the

strong base, sodium hydroxide (NaOH), is necessary to neutralize 1 mole of the weak acid, acetic

acid (CH3CO2H), as indicated in equation 3.

NaOH (aq) + CH3CO2H(aq) NaCH3CO2 (aq) + H2O (l) (Equation 1-3)

The sudden change in the pH of the solution shows that the titration has reached the equivalence

point. pH in an aqueous solution is related to its hydrogen ion concentration. Symbolically, the

hydrogen ion concentration is written as [H3O+]. pH is defined as the negative of the logarithm

of the hydrogen ion concentration.

pH = - log [H30+] (Equation 1-4)

pH scale is a method of expressing the acidity or basicity of a solution. Solutions having a pH <

7 are acidic, pH = 7 are neutral, pH > 7 are basic. For example, a solution having [H30+]= 2.35 x

10-2

M would have a pH of 1.629 and is acidic. Ph electrode will be used in this experiment. The

titration is initiated by inserting a pH electrode into a beaker containing the acid solution (pH

within 3-5). As sodium hydroxide, NaOH, is incrementally added to the acid solution, some of

the hydrogen ions will be neutralized. As the hydrogen ion concentration decreases, the pH of

the solution will gradually increase. When sufficient NaOH is added to completely neutralize the

acid (most of the H3O+ ions are removed from the solution), the next drop of NaOH added will

cause a sudden sharp increase in pH (figure 2-2). The volume of based required to completely

neutralized the acid is determine at the equivalence point of titration.

Figure 1-2: Acid-base titration curve of weak acid titrated with NaOH.

In this experiment, titration of vinegar sample with a standardized sodium hydroxide solution

will be done. To standardize the sodium hydroxide solution, of a primary standard acid solution

is initially prepared. In general, primary standard solutions are produce by dissolving a weighed

quantity of pure acid or base in a known volume of solution. Primary standard acid or bases have

several common characteristics:

they must be available in at least 99.9 purity

they must have a high molar mass to minimize error in weighing

they must be stable upon heating

they must be soluble in the solvent of interest

Potassium hydrogen phthalate KHC8H4O4, and oxalic acid, (COOH)2, are common primary

standard acid. Sodium carbonate, Na2CO3, is the most commonly used based. Most acids and

bases (e.g. HCl, CH3COOH, NaOH, and KOH) are most available as primary standard. To

standardize one of these acid or based solutions, titration of the solution with a primary standard

should be done. In this experiment, NaOH solution will be titrated with potassium hydrogen

phthalate (KHP). The equation for this reaction will be:

KHC8H4O4 (aq) + NaOH (aq) KNaC8H4O4 (aq) + H2O (l) (Equation 1-5)

Once the sodium hydroxide solution has beeb standardizes, it will be used to titrate 10.00mL

aliquots of vinegar. The equation for the reaction of vinegar with NaOH is

CH3COOH(aq) + NaOH(aq) NaCH3COO (aq) + H2O(l) (Equation 1-6)

Knowing the standardized NaOH concentration and using equation 6, we can determine the

molarity and percent by mass of acetic acid in the vinegar solution.

Sample calculation for standardizing a based with KHP

Figure 3 depicts the titration curve of 1.523 grams of KHP dissolved in20.0mL of distilled water

titrated with NaOH. Determine the molarity of the NaOH solution.

Figure 1-3: titration curve of KHP with NaOH. The volume of NaOH used at the equivalence point is 15.3 mL of NaOH.

Calculate the moles of KHP used in the titration.

8 4 48 4 4 8 4 4

8 4 4

11.523 0.007458

204.2

mol KHC H Og KHC H O x mol KHC H O

g KHC H O

From equation 5, calculate the moles of NaOH required neutralizing the moles of KHP.

10.007458 0.007458

1

mol NaOHmol KHP x mol NaOH

mol KHP

Calculate the molarity of the NaOH solution.

115.30 0.01530

1000

LmL NaOH x L NaOH

mL

0.007458 0.048750.4875

0.01530

mol NaOH mol NaOH mol NaOHM M NaOH

Lof solution L solution L solution

Sample calculations for determining the acetic acid concentration in vinegar by titration with

standard base

A 10.00 mL aliquot of vinegar requires 16.95 mL of the 0.4875 M standardized NaOH solution

to reach the equivalence point of the titration. Calculate the molarity and the percent by mass of

CH3COOH in the solution. Assume the density of the vinegar solution is 1.00g/mL.

Calculate the moles of NaOH that reacted.

116.95 0.01695

1000

LmL NaOH x L NaOH

mL

0.48750.01695 0.008263

1

mol NaOHL NaOH x mol NaOH

L NaOH solution

Calculate the moles of CH3COOH neutralized by the moles of NaOH

310.008263 0.008263

1

mol CH COOHmol NaOH x mol NaOH

mol NaOH

Calculate the molarity of the CH3COOH solution

3 31

10 0.0101000

LmLCH COOH x LCH COOH solution

mL

3 3 33

0.008263 0.82630.8263

0.01

mol CH COOH mol CH COOH mol CH COOHM M CH COOH

Lof solution L solution L solution

Calculate the mass of acetic acid in the solution

3 31

10 0.0101000

LmLCH COOH x LCH COOH solution

mL

3 33 3

3

0.8263 60.060.01 0.4963

1 1

mol CH COOH g CH COOHLCH COOH x x g CH COOH

L solution mol CH COOH

Calculate the mass of the acetic acid solution

33 3

3

110 10.00

1

g CH COOH solutionmLCH COOH solution x g CH COOH solution

mLCH COOH solution

Calculate the percent by mass of acetic acid in the solution

720

731

730

33

3

33 3

3

100%

0.4693100% 1.963%

10.00

g CH COOHpercent massCH COOH x

g CH COOH solution

g CH COOHpercent massCH COOH x CH COOH

g CH COOH

1.4 Procedure

1.4.1 Standardization of sodium hydroxide solution

1. Prepare 250 mL of approximately 0.6 M sodium hydroxide solution from NaOH

solid. The solution can be prepared in a beaker, and check the calculation with the

laboratory instructor prior to preparing the solution. Record your calculation.

2. Weight a 250 mL beaker and record the mass to the nearest 0.001g. Add 1.5 grams

of KHP to the beaker. Record the mass of the beaker and KHP to the nearest 0.001 g.

calculates the mass of KHP by difference and record the data. Add 30mL of distilled

water to the beaker. Stir the solution until the KHP has dissolved completely.

3. titrate this solution with NaOH and record the pH with 1 ml additions of NaOH

solution.

4. Repeat steps 1 to 3 to perform a second and third trial to standardize the NaOH

solution.

4. Plots the graph of pH versus NaOH. From the plots, determine the volume of

NaOH required neutralizing the KHP solution in each titration.

6. Calculate the molarity of sodium hydroxide for titration 1 and 2.

7. Calculate the average morality of sodium hydroxide solution for titration 1 and 2.

The resulting sodium hydroxide concentration will be used in part B of the

experiment.

1.4.2 Molarity of acetic acid and percent of vinegar

1. Transfer 10.00mL of vinegar to a clean, dry 250 mL beaker using a 10mL

volumetric pipette. . Add sufficient water, 75 to 100 mL, to cover the pH

electrode tip during the titration.

2. Add 1 ml of NaOH to the vinegar solution and record the pH

3. Repeat the above steps twice more

4. Plot the graph of pH vs volume NaOH added. And from the plots determine the

volume of NaOH required to neutralized vinegar in each titration. Record your data.

5. Calculate the molarity of acetic acid in vinegar for titration 1 and 2.

5. Calculate the average molarity of acetic acid fir each titration.

6. Calculate the percent by mass of acetic acid in vinegar for titration 1 and 2

7. Calculate the percent by mass of acetic acid in vinegar.

1.5 Result & Calculations

1.5.1 Standardization of sodium hydroxide solution

1. Calculations for preparing 150mL of approximately 0.6M sodium hydroxide

solution.

2.

Titration 1 Titration 2

Mass of beaker (g)

Mass of beaker + KHP (g)

Mass of KHP (g)

Volume of NaOH to neutralize

the KHP solution (mL)

3. Calculate the molarity of sodium hydroxide for each titration 1 and 2.

4. Calculate the average molarity of sodium hydroxide for each titration 1 and 2.

1.5.2 Standardization of sodium hydroxide solution

1.

Titration 1 Titration 1

Volume of

NaOH required

to neutralize

vinegar

2. Calculate the molarity of acetic acid in vinegar for titration 1 and 2.

3. Calculate the average molarity of acetic acid for each titration.

4. Calculate the % by mass of acetic acid in vinegar for titration 1 and 2.

5. Calculate the average percent by mass of acetic acid in vinegar.

EXPERIMENT 2: DETERMINATION OF THE Ka VALUE OF A WEAK ACID

The relative acidity of a substance or a system is important in many situations, such as in the

quality of drinking water, food preservation, soil conditions for agriculture and physiological

functions. The strength of an acid is measured based on its ability to donate protons to base.

The acid ionization constant, Ka, is a quantitative measure of the strength of an acid. The Ka

value is a characteristic of an acid and can be used to identify an unknown acid. The Ka value

indicates the relative strength of an acid. The larger the Ka value, the stronger the acid and vice

versa.

You are given 10 ml of two unknown acid solutions and required to determine the acid ionization

constants, Ka of weak acid solutions by titration with 0.1 M sodium hydroxide, NaOH and by

measuring the pH of the weak acid. You are also required to identify the unknown weak acid

solutions from the calculated Ka values obtained during the experiment.

Based on the actual identity of the unknown weak acid solutions, you are required to evaluate

and compare which method is more accurate.

EXPERIMENT 3: DETERMINATION OF CHROMIUM (VI) CONCENTRATIONS VIA

ABSORPTION SPECTROSCOPY

Many heavy metals, including chromium (Cr), are toxic even at low aqueous solution

concentrations. Chromium ions, Cr (III) or Cr (VI), are found naturally in rivers, lakes and

streams. Trivalent Cr (III) compounds are not usually considered as health hazards, but

hexavalent Cr (VI) compounds can be toxic if ingested or inhaled and have been established as

carcinogens.

You are required to determine the quantity of Cr (VI) present in a “polluted” water sample using

a spectrophotometer and a set of standard solutions. From your results you will conclude

whether the water sample is suitable for drinking and/or agricultural purposes.

EXPERIMENT 4: BASIC WATER PROPERTIES 1

Oxygen saturation or dissolved oxygen (DO) is a relative measure of the amount of oxygen

dissolved in a medium. DO is naturally present in water and tends to be manipulated in

order to suit spesific applications; in drinking water a higher DO level improves the taste, but

at the expense of higher rates of pipe corrosion, consequently industry tends to minimise the

DO level to reduce maintenance costs.

You are required to determine the DO level in a series of water samples and ascertain

whether they comply with Malaysian Water Standards.

EXPERIMENT 5: BASIC WATER PROPERTIES II

There are many chemicals present in water, not all of them are desirable. Chlorine is added

to water to kill bacteria which is good, but excessive quantities are hazardous to human

health. Iron from the degradation of pipes is inevitable and although its presence is not

hazardous to health, it does affect the taste of water and may discolour foods, which come in

contact with it. The presence of sulphates and phosphorous in water are a consequence of

using everyday items like cleaning products, which can pollute water supplies poisoning plant

and animal life.

You are required to determine the levels of chlorine (total and free), iron, sulphates and

phosphorous in a series of water samples and ascertain whether they comply with Malaysian

Water Standards.

2. LAB 6

SOAPS AND DETERGENT

2.1 Introduction

Soap is a generic term for the sodium or potassium salts of long-chain organic acids (fatty acids)

made from naturally occurring esters in animal fats and vegetable oils. All Organic acids contain

the RC02H functional group, where R is a shorthand notation for methyl, CH3-, ethyl CH3CH2-,

Propyl, CH3CH2CH2-, or more complex hydrocarbon chains called alkyl groups. Chemists use

the R shorthand notation because these groups can be very large and the hydrocarbon chain has

little effect on the compound's chemical reactivity. All esters contain the RC02R functional

group.

The R groups in soaps are hydrocarbon chains that generally contain 12 to18 carbon atoms.

Sodium fatty acids such as lauric (vegetable oil), palmitic (palm oil), and stearic (animal fat)

acids are just a few examples of soaps.

CH3(CH2)10COONa sodium laurate

CH3(CH2)16COONa sodium stearate

The hydrocarbon chain in soaps may contain saturated (no double bond) or unsaturated chains

(contains double bonds). Sodium salts are usually solid therefore; most bars of soap are of

sodium salts. Potassium salts are the basis of liquid soaps, shaving creams, and greases. Fats and

vegetable oils are triglycerides. Triglycerides in an ester derived from three fatty acids. A

triglyceride made from three lauric acid molecules is shown in Figure 6-1.

Saponification is the basic hydrolysis of an ester producing a carboxylic acid salt and an alcohol

(Eq.3-1) .A lone pair of electrons on the OH- is attracted to the partially positively charged C

atom in the C=O bond in the ester (Eq.6-1). The C-OR' bond breaks generating a carboxylic acid

(RC02H) and an alcohol (R'OH). In the presence of NaOH, carboxylic are converted to their

sodium salts (RCO2-Na

+).

When a triglyceride is saponified, three fatty acid salts (soaps) and glycerol are produced as

shown in Equation 6-2. The R groups in the triglyceride may or may not have the same

chailength (same number of carbons).Thus, different types of soaps may be produced from the

saponification of a particular triglyceride.

Figure 6-1: A Triglyceride molecule made from lauric acid and glycerol

(Equation 6-1)

(Equation 6-2)

2.2 Objectives

To :

(a) Prepare soap and compare its properties to that of a synthetics detergent.

2.3 Theory

Soap is the salt of a weak acid. Most organic acids arc weak acids. Consequently, hydrolysis

occurs to some extent when soap dissolves in water. Soap solutions tend to be slightly alkaline

(basic) due to partial hydrolysis of the acid (Eq. 3).

(Equation 6-3)

The cleansing action of soaps results from two effects. Soaps are wetting agents that reduce the

surface tension of water, allowing the water molecules to encounter the dirty object. They are also

emulsifying agents. "Dirt" frequently consists of a grease or oil along with other organic species.

In general, organic compounds are nonpolar. Water is a polar species. These two substances will

not dissolve in each other because of their dissimilar characteristics (the "Like Dissolves Like"

rule). Soaps cross the boundary between polar and nonpolar because they contain a polar

hydrophobic (water- hating) end and a polar hydrophilic (water loving) end as shown in Figure

6-2.

Figure 6-2: Molecular structure, a) a line drawing, b) of sodium stearate. In a line drawing, all carbon and hydrogen atoms are omitted at the intersection of each line as a shorthand method of drawing molecule. It is

understood that the C and H atoms are part of the molecule.

Because soaps have both polar and nonpolar region in the molecule, they are soluble in both

polar and nonpolar species. The hydrophobic (nonpolar) portion of soap is soluble in non polar

compound like grease and oils. The hydrophilic (polar) end dissolves in water. Soap molecules

surround the grease and oils and break them up into microscopic droplets can remain suspended

in the water. These suspended microscopic droplets are called micelles (Figure 6-3). Micelles

contain very small amounts of oil or grease in their center. Thus the oil or grease has been

dissolved in water forming an emulsion, one form of a suspension in water.

Figure 6-3: Formation of micelle

Water supplies in certain areas are acidic as a result of acid rain or pollution, or "hard" due to the

dissolved mineral content. Both acidic and "hard" water reduce the cleansing action of soap.

Soap is the salt of a weak acid. In the presence of a stronger acid, the sodium salt is converted to

an insoluble organic acid (Eq. 3-4).

(Equation 6-4)

"Hard water" contains dissolved Ca2+

, Mg2+

, and Fe 3+

, ions from the minerals that the water

passes over. Normally, soaps made from sodium and potassium fatty acid salts are soluble in

water. However, in the presence of these metal ions, the Na+ and K

+ soluble salts convert to

insoluble Ca2+

, Mg2+

, and Fe 3+

salts (E q. 3-5).

(Equation 6-5)

In either acidic or "hard" water, the soluble soaps form insoluble salts that becomes a scummy

ring on bathtubs and black areas on shirt collars .The cleansing ability of soap is reduced because

soap molecules are removed from solution. There are several techniques used to circumvent the

problems generated by hard water. Water can be "softened" via removing hard water ions from

solution using ion exchange techniques or by adding water-softening agents, such as sodium

phosphate (Na3PO4) o rsodium carbonate (Na2CO3). Water-softening agents react with the Ca2+

,

Mg2+

, and Fe 3+

, removing them from water (Eq. 6-6 and 6-7) and preventing the reaction of these

ions with soap (Eq. 6-4 and 6-5).

(Equation 6-6)

(Equation 6-7)

Thus “Syndets” was design to overcome the soap problem with “hard water”. Syndets differ

from soaps in that the nonpolar fatty acids groups are replaced with alkyl or aryl sulfonic acids

(ROS03H). The alkyl or aryl sulfonic acids have long chains of carbon atoms giving the

hydrophobic (nonpolar) end. The salt of the sulfonic acid (sulfonate) group forms the hydrophilic

end of the molecule. The difference in polar groups is one of the key distinctions between a soap

and a synthetic detergent. Syndets form micelles and cleanse in the same manner as soaps. Two

examples of synthetic detergents are shown in Figure 6-4.

Figure 2-4: Examples of synthetics detergents

Because sulfonic acid is a stronger acid than carboxylic acids, Syndets do not precipitate in

acidic solutions. Furthermore, alkyl and aryl sulfonates do not form insoluble salts in the

presence of the typical hard water ions. Thus, synthetic detergents remain soluble in both acidic

and "hard" water.

2.4 Procedure

2.4.1 Soap preparation

1. Place 25 mL of vegetable oil in a 250-mL Erlenmeyer flask. Add 20 mL of ethanol

and 25 mL of 6 M sodium hydroxide solution to the flask. Stir the mixture with d

stirring rod to mix the contents of the flask.

2. Heat the 250-mL flask in a boiling-water bath inside of a 600-mL beaker.

3. Stir the mixture continuously during the heating process to prevent the mixture

from foaming. If the mixture should foam to the point of nearly overflowing, remove

the flask from the boiling-water bath until the foaming subsides, then continue

heating. Heat the mixture for 20-30 minute or until the alcohol odor is no longer

detectable.

4. Remove the paste-like mixture from the boiling-water bath and cool the flask in an

ice bath for 10-15 minutes.

5. While the flask is cooling assemble the vacuum filtration apparatus shown in

Figure 3-5. Secure the vacuum flask to a ring stand with a utility clamp to pre-vent

the apparatus from toppling over.

Figure 6-5: Vacuum Filtration apparatus

6. Weigh a piece of filter paper to the nearest 0.001 g and record the mass. Place the

filter paper inside the Buchner funnel. Moisten the paper with water so that it fits

flush in the bottom of the funnel.

7. Once the flask has cooled, add 150 mL of saturated sodium chloride (NaCl) solution

to the flask to "salt out" the soap.

8. Slowly turn on the water at the aspirator. Pour the mixture from the flask into the

Buchner funnel. Once all of the liquid has filtered through the funnel, wash the soap

with 10 mL of ice-cold water. Continue the suction filtration until a11 of the water is

removed from the soap.

9. Remove the soap from the funnel and press it between two paper towels to dry it.

Weigh the filter paper and dried soap, and record the mass to the nearest 0.001 and

determine the mass of the soap by difference and record the mass.

2.4.2 Comparison of soap and detergent properties- precipitation and emulsification

1. Prepare a stock soap solution by dissolving 2g of your prepared soap in 100 mL of

boiling, distilled water. Stir the mixture until the soap has dissolved and allow the

solution to cool.

2. Repeat step 1 using 2 g of synthetic detergent (e.g., Dynamo). When both solutions

are cool, determine the pH of each solution using pH paper.

3. Label three test tubes as test tube 1, 2, and 3. Add 4 drops of mineral oil to each

test tube. Add -5 mL of distilled water to test tube 1. Add -5 mL of stock soap

solution to test tube 2. Add -5 mL of stock synthetic detergent to test tube 3.

4. Mix each solution by shaking and let stand for three to five minutes. Note which of

the solutions, if any, emulsifies the oil by forming a single layer.

5. Pour the mixtures into the Waste Container. Clean and dry the three test tubes.

6. Label three test tubes as test tube 1, 2, and 3. Place 2 mL of stock soap solution in

each of the three test tubes. Add 2 mL of 1% CaCl2 solution to test tube 1. Add 2 mL

of 1% MgC12 solution to test tube 2. Add 2 mL of l% FeCl2 solution to test tube 3.

Shake each test tube to mix the solutions. Record your observation.

7. Add 4 drops of mineral oil to each of the test tubes in Step 6. Shake each test tube to

mix the solutions and let the solutions stand for three five minutes. Note which of the

solutions, if any, emulsifies the oil by forming a single layer.

8. Repeat Steps 6-7 using -2 mL of stock detergent solution. Which solutions form a

precipitate?

9. Note which of the solutions, if any, emulsifies the oil by forming a single layer.

10. Pour the mixtures into the Waste Container. Clean and dry the test tube.

11. Place 5 ml, of stock soap solution in cine clean test tube and 5 rnL of stock

detergent solution in a second test tube. Add 1 M HC1 one drop at a time to both

solutions until the pH in each test tube is equal to 3. (Use pH paper to

measure).Count the number of drops of acid added to each mixture. Does a precipitate

form in either mixture?

12. Add 1 drops of mineral oil to each test tube in Step 11. Shake each test tube to

mix the solution. Is the oil emulsified in either mixture?.

2.4.3 Comparison of the cleaning abilities of a soap and detergent.

1. Clean, dry, and label three beakers. Place 20 mL of stock soap solution (from Step

1 in section 3.4.3) in the 1st beaker. Place 20 mL of stock detergent solution (from

Step 2 in section 6.4.4) in the 2nd beaker. Place 20 mL of a commercial liquid

Dynamo.

2. Obtain three cloth test strips that have been soaked in tomato sauce and place one

strip in each of the beakers. Place one cloth strip in beaker 1 (from above), one cloth

strip in beaker 2, and one cloth strip in beaker 3. Repeatedly stir each solution with a

stirring rod for 5 minutes.

3. Remove the cloth strips from the soap and detergent solution and squeeze out the

excess water. Visually compare each cloth strip to determine their relative cleanliness.

Record your observations.

2.5 Result & Calculations

2.5.1 Soap preparation

Mass of Filter (g)

Mass of filter paper

+ soap (g)

Mass of soap

recovered (g)

2.5.2 Comparison of soap and detergent properties.

Brand name of

synthetics

detergent

pH of soap

solution

pH of synthetics

detergent

solution

Answer yes or no in the space provided below

System Emulsification

Occurred

Distilled water

Soap

detergent

Hard and acidic

2.5.3 Cleansing comparison of a soap and detergents

Visually compare each cloth strip to determine their relative cleanliness. Record your

observations.

Compare the cleansing ability of the two detergents.

System Precipitate Oil emulsified

Soap x Soap x

CaCl2

MgCl 2

FeCl3

Acidic