The measurement of the strength of an acid by the quantity of alkali required to neutralise it (termed acidimetry), and the measurement of the strength of an alkali by the quantity of acid required to neutralise it (termed alkalimetry), will be considered in this chapter. The student is asked to determine volumetrically:
THE ESTIMATION OF CHLORINE IN BaCl2,2H2O VOLUMETRICALLY BY STANDARD ALKALI AND ACID.
Apparatus and Reagents.—The usual volumetric outfit. For practice, the student may procure some pure barium chloride. The standard N/10 Na2CO3 and N/10 H2SO4 are again made use of.
Method, Reactions.—To the BaCl2 solution N/10 Na2CO3 is added from the burette till decidedly alkaline.
BaCl2 + Na2CO3 + xNa2CO3 = BaCO3 + NaCl + xNa2CO3
The total volume of alkali added is noted. Then, when the precipitate has settled, a known volume of the clear liquid is taken and the excess of alkali (xNa2CO3) is titrated with N/10 H2SO4, From the figures thus obtained the amount of alkali that combined with the acid of the barium chloride is obtained, and from this the percentage of chlorine present.
Details of the Analysis.—Weigh out two portions of about .3 gm. of the pure dry salt. Transfer to 200 c.c. beakers, and dissolve each portion in 150 c.cs. distilled water. When dissolved, add from a burette N/10 Na2CO3 till decidedly alkaline to litmus paper (about 35 c.cs, should be sufficient for .3 gm. of the salt). Transfer the liquid and precipitate to the 250 c.c. flask, washing out the beaker and bringing the solution up to the mark.
When the precipitate has settled, remove by the 50 c.c. pipette 50 c.cs. of the clear solution.
Titrate this as usual with N/10 H2SO4. Remove another 50 c.cs. and repeat the titration. Submit the second .3 gm. portion to the same treatment.
Calculation of Results. — Assume as an example that .3 gm. of the salt are taken and that 35.1 c.cs. N/10 Na2CO3 were added, and that 2.1 c.cs. of N/10 H2SO4 were required in the titration of 50 c.cs. of the clear solution.
Then 5 x 2.1 = 10.5 c.cs. N/10 H2SO4 are required to neutralise the excess of alkali.
Therefore the amount of alkali required to combine with the acid of the salt is 35.1-10.5 = 24.6 c.cs. N/10 Na2CO3.
Now 1 c.c. N/10 Na2CO3 = .0035 gm. Cl, therefore
24.6 c.c. N/10 Na2CO3 = .0861 gm. Cl.
The percentage of chlorine = .0861 x 100/.3 = 28.7%
The theoretical percentage of chlorine in pure BaCl2 2H2O is 71 x 100/244 = 29.1.
On looking through these calcuations, the student should notice that any error in the reading of the acid used in titration is magnified five times in the multiplication. Thus, if instead of 2.1 c.cs, the reading be 2 c.cs., the results work out as follows:
Percentage chlorine = .8785 x 100/.3 = 29.28.
Great care is therefore necessary in reading the burette. The liability to error may be lessened by titrating two lots of 100 c.cs. each, or better still, one lot of 200 c.cs. of the clear liquid. Any error in reading thus suffers much less magnification by multiplication.
Note.—Store the remainder of the standard solutions in Winchester bottles.