For the preparation of solutions of molar and normal concentrations, a sample of the substance is weighed on an analytical balance, and the solutions are prepared in a volumetric flask. When preparing solutions of acids, the desired volume of the concentrated acid solution is measured by a burette with a glass faucet.
The weighed portion of the solute is calculated to four decimal places, and the molecular weights are taken with the accuracy with which they are given in the reference tables. The volume of the concentrated acid is calculated to the second decimal place.
Example 1. How many grams of barium chloride are needed to prepare 2 liters of 0.2 M solution?
Decision. The molecular weight of barium chloride is 208.27. Therefore 1 l of 0.2 M solution should contain 208.27-0.2 = = 41.654 g of BaC12. For the preparation of 2 liters, 41.654-2 = 83.308 g of BaCl2 will be required.
Example 2. How many grams of anhydrous soda Na2CO3 will be needed to prepare 500 ml 0.1 n. solution?
Decision. The molecular weight of soda is 106.004; equivalent unit weight 5Na2C03 = M: 2 = 53.002; 0.1 eq. = 5,3002 g.
1000 ml 0.1 n. solutions contain 5,3002 g of Na2C03500 “” “” ” x “Na2C03
5,3002-500x = —— Goooo—- = 2-6501 g Na2C03.
Example 3. How much concentrated sulfuric acid (96%: d = l, 84) is required for the preparation of 2 liters of 0.05 n. sulfuric acid solution?
Decision. The molecular weight of sulfuric acid is 98.08. Equivalent mass of sulfuric acid 3h2so4 = M: 2 = 98.08: 2 = 49.04 g. Mass 0.05 eq. = 49.04-0.05 = 2.452
We find how much H2SO4 should be contained in 2 liters of 0.05 n. solution:
In order to determine how much it is necessary to take 96% of H2S04 solution, we make up the proportion:
We recalculate this quantity by volume:,. P 5.11
K = 7 = TJ = 2 ‘77 ml –
Thus, for the preparation of 2 liters of 0.05 n. solution should take 2.77 ml of concentrated sulfuric acid.
Example 4. Calculate the titer of NaOH solution, if it is known that its exact concentration is 0.0520 n.
Decision. Recall that the title is the content in 1 ml of a solution of a substance in grams. Equivalent mass of NaOH = 40 01 g We find how many grams of NaOH are contained in 1 liter of this solution:
40.01-0.0520 = 2.0805 g.
1 titer of solution: -sch = – = 0,00208 g / ml. You can also use the formula:
where T is the titer, g / ml; E is the equivalent mass; N is the normality of the solution.
Then the titer of this solution:
“NaOH = ——— jooo—— 0.00208 g / ml.
„“ Р ие Р 5 – Calculate the normal concentration of the HN03 solution, if it is known that the titer of this solution is 0.0065 To calculate, we use the formula:
The normal concentration of a solution of nitric acid is:
– V = 63.05 = 0.1030 n.
Example 6. What is the normal concentration of the solution, if it is known that 200 ml of this solution contains 2.6501 g of Na2C03
Decision. As calculated in Example 2, Zma2co (= 53.002. Finding how many equivalents is 2.6501 g Na2C03: D2.6501: 53.002 = 0.05 equiv. /
In order to calculate the normal concentration of the solution, we make up the proportion:
200 ml contain 0.05 eq.
1000 »» x ”
1000-0.05x =—————— = 0.25 eq.
1 l of this solution will contain 0.25 equivalents, i.e. the solution will be 0.25 n.
For such a calculation, you can use the formula:
Where R – amount of substance in grams; Uh – equivalent mass of substance; V – solution volume in milliliters.
Zya2so3 = 53,002, then the normal concentration of this solution