FREE ALKALI OR ACID.
(a) Alcoholic Method.
Test a freshly cut surface of the soap with a few drops of an alcoholic phenolphthalein solution. If it does not turn red it may be assumed free fat is present; should a red color appear, free alkali is present. In any case dissolve 2 to 5 grams of soap in 100 cubic centimeters of neutralized alcohol and heat to boiling until in solution. Filter off the undissolved portion containing carbonate, etc., and wash with alcohol. Add phenolphthalein to the filtrate and titrate with N/10 acid and calculate the per cent. of free alkali as sodium or potassium hydroxide. Should the filtrate be acid instead of alkaline, titrate with N/10 alkali and calculate the percentage of free fatty acid as oleic acid.
The insoluble portion remaining on the filter paper is washed with water until all the carbonate is dissolved. The washings are then titrated with N/10 sulfuric acid and expressed as sodium or potassium carbonate. Should borates or silicates be present it is possible to express in terms of these. If borax is present the carbon dioxide is boiled off after neutralizing exactly to methyl orange; cool, add mannite and phenolphthalein and titrate the boric acid with standard alkali.
(b) Bosshard and Huggenberg Method.[14]
In using the alcoholic method for the determination of the free alkali or fat in soap there is a possibility of both free fat and free alkali being present. Upon boiling in an alcoholic solution the fat will be saponified, thus introducing an error in the analysis. The method of Bosshard and Huggenberg overcomes this objection. Their method is briefly as follows:
Reagents.
1. N/10 hydrochloric acid to standardize N/10 alcoholic sodium hydroxide.
2. Approximately N/10 alcoholic sodium hydroxide to fix and control the N/40 stearic acid.
3. N/40 stearic acid. Preparation: About 7.1 grams of stearic acid are dissolved in one liter of absolute alcohol, the solution filtered, the strength determined by titration against N/10 NaOH and then protected in a well stoppered bottle, or better still connected directly to the burette.
4. A 10 per cent. solution of barium chloride. Preparation: 100 grams of barium chloride are dissolved in one liter of distilled water and filtered. The neutrality of the solution should be proven as it must be neutral.
5. α naphtholphthalein indicator according to Sorenson. Preparation: 0.1 gram of α naphtholphthalein is dissolved in 150 cubic centimeters of alcohol and 100 cubic centimeters of water. For every 10 cubic centimeters of liquid use at least 12 drops of indicator.
6. Phenolphthalein solution 1 gram to 100 cubic centimeter 96 per cent. alcohol.
7. Solvent, 50 per cent. alcohol neutralized.
MANIPULATION.
First—Determine the strength of the N/10 alcoholic sodium hydroxide in terms of N/10 hydrochloric acid and calculate the factor, e. g.:
| 10 c.c. N/10 alcoholic NaOH | = 9.95 N/10 HCl} | 9.96 |
| 10 c.c. N/10 alcoholic NaOH | = 9.96 N/10 HCl} |
The alcoholic N/10 NaOH has a factor of 0.996.
Second—Control the N/40 stearic acid with the above alkali to obtain its factor, e. g.:
| 40 c.c. N/40 alcoholic stearic acid = | 10.18 c.c. N/10 NaOH } | } 10.2 |
| 40 c.c. N/40 alcoholic stearic acid = | 10.22 c.c. N/10 NaOH } |
10.2 × F N/10 NaOH (0.996) = Factor N/40 stearic acid
∴Factor N/40 stearic acid = 1.016.
Third—About 5 grams of soap are weighed and dissolved in 100 cubic centimeters of 50 per cent. neutralized alcohol in a 250 cubic centimeter Erlenmeyer flask over a water bath and connected with a reflux condensor. When completely dissolved, which takes but a few moments, it is cooled by allowing a stream of running water to run over the outside of the flask.
Fourth—The soap is precipitated with 15 to 20 cubic centimeters of the 10 per cent. barium chloride solution.
Fifth—After the addition of 2 to 5 cubic centimeters of α naphtholphthalein solution the solution is titrated with N/40 alcoholic stearic acid. α naphtholphthalein is red with an excess of stearic acid. To mark the color changes it is advisable to first run a few blanks until the eye has become accustomed to the change in the indicator in the same way. The change from green to red can then be carefully observed.
Let us presume 5 grams of soap were taken for the analysis and 20 cubic centimeters of N/40 stearic acid were required for the titration then to calculate the amount of NaOH since the stearic factor is 1.016.
20 × 1.016 = 20.32 N/40 stearic acid really required.
1 cubic centimeter N/40 stearic acid = 0.02 per cent. NaOH for 5 grams soap.
Δ 20.32 cubic centimeters N/40 stearic acid = 0.02 × 20.32 per cent. NaOH for 5 grams soap.
Hence the soap contains 0.4064 per cent. NaOH.
It is necessary, however, to make a correction by this method. When the free alkali amounts to over 0.1 per cent. the correction is + 0.01, and when the free alkali exceeds 0.4 per cent. the correction is + 0.04, hence in the above case we multiply 0.004064 by 0.04, add this amount to 0.004064 and multiply by 100 to obtain the true percentage. Should the alkalinity have been near 0.1 per cent. we would have multiplied by 0.01 and added this.
If carbonate is also present in the soap, another 5 grams of soap is dissolved in 100 cubic centimeters of 50 per cent. alcohol and the solution titrated directly after cooling with N/40 stearic acid, using α naphtholphthalein or phenolphthalein as an indicator, without the addition of barium chloride. From the difference of the two titrations the alkali present as carbonate is determined.
If the decomposed soap solution is colorless with phenolphthalein, free fatty acids are present, which may be quickly determined with alcoholic N/10 sodium hydroxide.