Acids like hydrochloric acid (HCl), sulphuric acid (H_{2}SO_{4}), nitric acid (HN0_{3}) and acetic acid (CH_{3}COOH) which are commonly used in laboratory are originally available as concentrated aqueous solutions. The solutions required in laboratory are prepared by diluting the concentrated solutions with water. Knowing the concentration of the concentrated acid we can calculate how much volume of this acid must be diluted to prepare a definite volume of dilute acid of certain required concentration. The approximate concentrations of some acids as commercially available are given in Table 11.2.

**Table 11.2. Approximate Concentrations of Some Acids**

Concentrated Acid Approximate Approximate

Normality (N) Molarity (M)

Hydrochloric acid (HCI) 12 12

Sulphuric acid (H_{2}SO_{4}) 36 18

Nitric acid (HN0_{3}) 16 16

Glacial acetic acid 17 17

(CH_{3}FOOH)

**STOCK OR COMMERCIAL SOLUTIONS**

A stock solution is a concentrated solution (standard or non-standard) whose concentration is known, which can be stored and diluted as required to give solutions of lower concentration. A stock solution should be chemically stable, so that it doesn’t change with storage. Keeping suitable quantities of stock solution can save time during term in making up solution, as they only need dilution. Thus 1 dm3 of a 10M solution, for example, would make 10 dm3 of 1M solution when diluted. Concentrated acids are one example of a stock solution, which can be diluted as required. When making up bench reagents the stock solution can be measured out with a measuring cylinder; for analytical solutions it should be measured out using a pipette, depending on how precisely the stock solution is known. If the stock solution is only approximate, then it is best to dilute without taking to much care and then standardise the final solution. The exact final concentration is then known.

**DILUTION FACTOR**

We often need to dilute one solution, whose concentration is known, to give another of known concentration. Although this is a simple task some students find it difficult. The basic idea to remember is that when we dilute a solution the amount of solute stays the same, although its concentration decreases.

In any solution:moles of solute

=volume (dm3) x concentration (moVdm3)

=volume (cm3)/1000 x concentration (moVdm3)

Diluting with water keeps the number of moles constant i.e., the product of volume and concentration is constant.

Thus:

volume (i) x concentration (i)

= volume (f) x concentration (f)

Here, i = initial; f = final

V _{i} x M _{i} = V _{f} x M _{f}

Thus, the above relation can be used to prepare solutions of various lower concentrations by calculating the volume of water to be added to the more concentrated solution. The number of times the volume of the more concentrated solution is diluted to obtain the less concentrated solution is called the dilution factor. The above idea is explained by the following example.

**STEPS INVOLVED IN THE PREPARATION OF SOLUTIONS FROM LIQUID SOLUTES**

** **Preparation of250 cm_{3} of 5 M HCl from the given 12M HCl.

To prepare 250 cm_{3} of 5 M HCl from 12 M HCl, the volume of concentrated acid required ·can be calculated by using the molarity equation:

M_{1} V_{1} = M_{2} V_{2}

where, M_{1 }= Molarity of concentrated acid= 12M

V_{I }= Volume of concentrated acid required = ?

M_{2} = Molarity of dilute acid to be prepared = 5M

V _{2} = Volume of dilute acid to be prepared = 250 cm3.

12 x V _{1 }= 5 x 250

V_{1 }= 5 x 250 / 12 = 104.2 cm_{2}

Therefore, 104.2 cm3 of concentrated acid should be taken and diluted with water to make the volume 250 cm3. The resulting solution will be 5 M HCl.

The apparatus required is a measuring cylinder, 250 cm_{3} measuring flask, beakers and glass rod. Procedure

1. Calculate the volume of ,12 M HCl required for preparing 250 cm_{3} of 5 M HCl.

2. With the help of a measuring cylinder, take about 100 cm_{3} of distilled water in a 400 cm3 beaker. (This volume of water is slightly less than the volume of water which will be required).

3. Take 104.2 cm_{3} of given concentrated HCl in a measuring cylinder. Add it slowly and carefully, with stirring with a glass rod, to the water taken in the beaker. Cool the beaker under tap water from time to time.

4. After all the acid has been added, transfer the solution from the beaker to a 250 cm_{3} measuring flask. Add more of water so that the total volume becomes 250 cm3.

5. Stopper the measuring flask and shake the solution well to make it uniform. Label it as 5 M HCl.