APPENDIX 453 



What may be considered as the valence of a base is shown by the number of 

 hydroxyls combined with it; that of an acid by the number of replaceable hydrogen 

 atoms which it contains. 



To make a normal solution, dissolve in distilled water a weight in grams equal 

 to the sum of the atomic weights of the substance divided by its valence, and make 

 up the volume to exactly 1000 c.c. 



NaOH is univalent. Na = 23. O = i6. H=i. Dissolve 40 grams NaOH in 

 water and make up to exactly 1000 c.c. 



Oxalic acid is COOH COOH+2H 2 O which gives it a molecular weight of 126. 

 As it contains two carboxyl groups it is dibasic, and it is necessary to divide the 

 molecular weight by 2, so that for a normal solution of oxalic acid we dissolve 63 

 grams in a volume of distilled water made up to 1000 c.c. 



If a chemical laboratory is not accessible one may prepare normal solutions with 

 an error so slight as to be unimportant in clinical work in the following way: 



Sodium hydrate being very hygroscopic, it is impossible to accurately prepare a 

 normal solution by directly weighing out the substance. Instead, select perfect 

 crystals of oxalic acid, such as can be obtained in a drug store, and weigh out on the 

 most accurate apothecary scales obtainable exactly 6.3 grams of the most perfect 

 crystals in the bottle. Put these preferably in a volumetric flask and make up with 

 distilled water to 1000 c.c. Less accurate is the use of a measuring cylinder. If 

 care is used this should give N/io solution of oxalic acid in which the error is less 

 than i%. 



Having N/io acid at hand, we may prepare N/io NaOH in the following way: 

 Weigh out an excess of sodium hydrate (5 grams of stick caustic soda) and dissolve 

 in 1 100 c.c. of distilled water. Take up 10 c.c. of this solution with a pipette and let 

 it run into a beaker. Add 6 drops of phenolphthalein solution. This gives a violet- 

 pink color. Fill the burette with the N/io oxalic acid solution and let it run into 

 the sodium hydrate solution in the beaker until the pink is just discharged. Reading 

 off the number of cubic centimeters of the N/io acid used, we know the strength 

 of the sodium hydrate solution. It is well to repeat the titration and take an 

 average. 



If 10.5 c.c. of the oxalic acid solution were required it would show that the sodium 

 hydrate solution was stronger than N/io, as only 10 c.c. would have been necessary 

 if the NaOH solution had been N/io. It is therefore necessary to dilute the sodium- 

 hydrate solution in the proportion of 10 to 10.5. Measure exactly 1000 c.c. of the 

 too concentrated sodium-hydrate solution and add to it 50 c.c. of distilled water, mix 

 thoroughly, and we have 1050 c.c. of N/io solution of NaOH. icooX 10.5 = 10,500. 

 1 0,500 -1-10 = 1050. 



As Acidum hydrochloricum U. S. P. is about two-thirds water (68.1%) to make 

 N/io HC1, which would require 3.65 in 1000 c.c., it would be necessary to take about 

 three times this amount of U. S. P.^cid. Take 12 c.c. of the acid and add distilled 

 water to make 1 100 c.c. Put 10 c.c. of this dilute solution in a beaker. Add phenol- 

 phthalein solution and titrate. If n c.c. of N/io NaOH were required it would be 

 necessary to add 100 c.c. of water to a volume of 1000 c.c. of the diluted hydrochloric 

 acid. ioooXii = n,ooo-T-io = iioo. 



Other acid and alkali solutions can be made as for N/io HC1 and N/io NaOH. 



