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Mol. wt. of Na₂ CO3 = 2 x 23 + 12 + 48 106

Equivalent wt.Mol. wt./2 =53

By the definition of a normal solution.

1000 ml of (N) Na₂ CO3 solution = 53 gm

1000 ml of N/ 10 Na₂ CO3 solution = 5.3 gm

1000 ml of N/10 Na₂ CO3 solution = 0.53 gm

Hence, 0.53 gms of pure dryNa₂ CO3 are required to prepare 100 ml of exactly decinormal solution.


(1) 100 ml measuring flask 

(2) Watch glass 

(3) Desiccator

(4) Porcelain basin 

(5) Funnel 

(6) Pipette.


  Na₂ CO3


i) Heat some pure anhydrous sodium carbonate in a porcelain basin for some time so as to expel all moisture, cool it and then put it in a desiccator.

ii) Accurately weigh out about 0.53 gm of Na2 CO3 in a chemical balance. A watch glass may be used for the purpose of weighing.

iii) Transfer it quantitatively into a clean 100 ml measuring flask with the help of a small funnel. Wash down the adhering particles with some distilled water (use a wash bottle). Gently swirl the flask for some time. 

iv) Dilute the solution with more distilled water, and swirl the flask thoroughly till the substance dissolves completely.

v) Add more distilled water carefully till the lower meniscus of the curved surface of the solution is up to the mark on the neck of the flask (better add the last portion of water dropwise from a pipette.)

iv) Close the flask by means of a stopper, and shake well so as to make the solution homogeneous.


Generally, It is difficult to weigh out exactly the required amount, though it is not impossible. Suppose the weights taken are as follows:

wt. of empty watch glass = 10.5 gm (say)

wt. of watch glass + Na₂ CO3 = 11.12 gm (say)

Weight of Na₂ CO3 taken= 0.62 gm

Now, 0.53 gm in 100 ml makes the solution N/10

0.62 gm in 100 ml makes =0.62/0.53 x N/10

    = 1.17 x N/10

    = N/10  (f = 1.17)

So our solution is not exactly decinormal but 1.17 times N/10. The coefficient 1.17 is called the factor of the strength of the solution prepared.

Hence, factor (f) = weight actually taken/wt required theoretically


The standard solution of Na₂ CO3 prepared above is called a primary standard. But all substances can not be weighed to make primary standards. For instance, NaOH, HCl, H₂SO4 ,HNO3 etc can not be used to make primary standards. For a substance to be useful as the primary standard, it should satisfy the following.

i) It must not react with or absorb the components of the atmosphere.

ii) It must have a high percentage purity.

iii) It should have a high mol. wt.

iv) It should be non-toxic.

          In case of NaOH, HCl, H₂SO4 or HNO3, approximately decinormal solutions are prepared first, and then these solutions are standardised by titration against the primary standard. These standardised solutions are called secondary standards.

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