Apparatus and Reagents.—The usual volumetric and gravimetric apparatus. The student may take for analysis a sample of the ordinary bench reagent 5E. HCl. He will also require about 10 gms. pure NaHCO3, and should test this substance for Cl and SO3. If these are found, place a small plug of cotton-wool in the neck of a funnel, and on it pack about 100 gms. of the bicarbonate, and wash repeatedly with small lots of distilled water to remove the chlorides and sulphates, which are more soluble than the bicarbonate. Test the washings, and when free from these impurities spread the remainder of the salt between several folds of stout filter paper, and dry in the water oven at a temperature of 60° C., and when dry transfer to a clean weighing bottle or tube.
Method Reactions.—A certain volume of HCl is taken at a certain temperature. Two drops of methyl orange solution are added, and the solution is ‘titrated’ (that is, its contents are ascertained) with a standard solution of Na2CO3.
2HCl + Na2CO3 = 2NaCl + H2O + CO2
The value of the Na2CO3 solution being known, and the volume used being also known, the strength of the HCl is easily calculated.
Preparation of the Standard Solution.—A ‘normal’ solution of Na2CO3 contains 46+ 12+ 48/2 = 53 gms. of the salt per litre. For the present purpose a N/10 solution containing 5.3 gms. Na2CO3 per litre will answer.
Weigh out about 9 gm. pure NaHCO3 and transfer to a porcelain crucible which has been ignited and weighed. Heat to dull redness for fifteen minutes, stirring occasionally with a stout platinum wire or glass rod. Cool in the desiccator and weigh, and it will be found that a little over 5.3 gms. Na2CO3, remain. Adjust quickly to 5.3 gms. by removing a little of the salt on the tip of a small spatula. Adjustment within .001 gm. will suffice. If the student cannot perform this adjustment quickly, moisture will be absorbed appreciably. In such a case it will be preferable to note the exact weight of Na2CO3, omit the adjustment, and allow for the difference by calculating the volume of solution. If, for example, 5.4521 gms. Na2CO3 are present, make up to 1000 x 5.4521/5.3 c.cs. instead of 1000 c.cs.
Empty the contents of the crucible into a 200 c.c. beaker and wash out the crucible with several small washes of hot water. Add more hot water till the volume of solution is about 100 c.cs. Stir carefully with a light glass rod till the salt is completely dissolved. Allow the solution to cool. Make up to 1000 c.cs. by transferring to the litre flask and washing out the beaker with distilled water Further water is added till, at a temperature of about 16° C., the solution reaches the mark. If the adjustment of Na2CO3 to 5.3 gms. was not made, add the additional amount of distilled water from the burette.
If correctly made up from pure Na2CO3, 1 c.c. of this solution should contain 5.3/1000 = .0053 gm. Na2CO3, which is equivalent to .00365 gm. HCl.
Checking the Standard Solution.—In order that the student may ascertain the true value of this solution he must check it; it may be correct or it may not. When once carefully checked this solution will serve for the preparation of standard acids or other standard alkalies. On the accuracy of this one standard depends the accuracy of all results.
Hydrochloric acid, the substance to be analysed, may also be used to check the solution as, follows:—Fill the 50 c.c. burette with E/5 HCl. prepared approximately from the reagent bottle. Into each of four narrow 200 c.c. beakers measure very carefully 20 c.cs. E/5 HCl. Cover and set aside two of these portions.
Fill a clean burette with N/10 Na2CO3. To the contents of each of the two beakers add 2 drops methyl orange solution, adjust the solution in the burette to zero, and under it, on a small square of white paper, place one of the beakers and contents. Proceed to titrate the solution by slowly adding the solution from the burette a little at a time, stirring with a light glass rod between each addition. About 36 c.cs. may thus be run in, but after this the additions must be made more cautiously, till finally the solution is added drop by drop till the pink colour just disappears. Note the number of c.cs. used. Repeat the operation with the second portion. The duplicates should not differ by more than .1 c.c. provided the burette from which the E/5 HCl was measured has been calibrated, and any corrections necessary made. The same remark applies to the burette used for the titration.
If the duplicates do not agree closely, weigh out another 20 c.cs. E/5 HCl and repeat the operation. If this result does not agree with one of the first two, either the burettes are badly calibrated or the student’s work is bad, or perhaps both. It is of course assumed that the temperatures of the two solutions are approximately equal.
Proceed now to estimate the chlorine in the two reserved samples by means of the Gravimetric method previously given, taking care that any considerable excess of AgNO3 is avoided. The duplicates should agree closely.
As a concrete example assume the following results:—
Average of titrations = 40.1 c.cs.
Average Cl in 20 c.cs. E/5 HCl=.1432 gm. = .1472 gm. HCl.
But 1 c.c. N/10 Na2CO3 should be equivalent to .0o365 gm, HCl.
Therefore .1472 gm, HCl should require .1472/.00365 = 40.33 c.c. N/10 Na2CO3.
That is, the N/10 Na2CO3 solution is slightly strong, 40.1 c.cs. having the neutralising power of 40.33 c.cs. of the desired solution. The bulk of the solution must now be diluted in the proportion of 40.1 to 40.33. Transfer the alkaline solution to a measuring cylinder. Note its volume; for example, 880 c.cs.
Now
401 : 103 : : 880 : 884.4
Add therefore to the alkali solution 4.4 c.cs. distilled water from a burette. The student now has checked the accuracy of and adjusted the strength of the standard solution to n/10 Na2CO3. It may appear to him at this stage that volumetric analysis can hardly lay claim to speed. He must remember, however, that from this one standard solution, solutions sufficient for hundreds of analysis may quickly and accurately be prepared. In practice it is usual to prepare instead of N/10 solutions a stock of N. alkali and N. acid ; but as the student has only a few analyses to perform, solutions will serve.
The actual Analysis.—The student may now proceed to find the strength of any of the common mineral acids. A bottle is, for example, labelled E. HCl. The student is asked to determine the accuracy of the label value.
Knowing the approximate strength of the acid, the student will see that a 5 c.c. sample will require approximately 50 c.cs. N/10 Na2CO3. As the error in measuring such a sample may be large in proportion to the volume measured, this error may be relatively lessened by measuring out 25 c.cs. into a 250 c.c. flask, and making up the bulk with distilled water to 250 c.cs. After shaking, remove 25 c.cs. with the pipette. This sample represents 1/10th of the total, that is, 2.5 c.cs.
To this sample add 2 drops of methyl orange and titrate with the N/10 Na2CO3. Repeat the titration on a fresh 25 c.cs. from the flask. Proceed till the results closely agree.
Now take another 25 c.cs. of the E. HCl; make up as before to 250 c.cs. and repeat the titrations. The results should agree with the former set.
Calculation.—As an example, assume that the average of the readings is 26.2 c.cs. N/10 Na2CO3. Then 2.5 c.cs. E, HCl are neutralised by 26.2 c.cs. N/10 Na2CO3. If accurately made up, they would be neutralised by 25 c.cs. N/10 Na2CO3. That is, the strength of the acid analysed is 26.2/25.0 E., or 1.048 E.
Or in percentages, 1 c.c. N/10 Na2CO3 =.00365 gm. HCl.
26.2 c.cs. N/10 Na2CO3 = 26.2x.00365 gm. HCl =.09563 gm. HCl.
If 2.5 c.cs. dilute HCl contain .09563 gm. HCl, then 100 c.cs. of the dilute acid contain=3.8252 gms. HCl. That is, the solution contains 3.8252% HCl.