[PDF] Preparation of Solutions- Normal solutions





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Laboratory Solution Preparation

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84-1201 Carolina’s Solution Preparation Manual

the final volume and concentration of solution desired and then determine the amount of solute Dilute the solute in sufficient solvent to produce the final volume of solution desired For example to prepare 100 mL of a 10 by volume solution of acetic acid dilute 10 mL acetic acid with distilled or deionized water to make 100 mL of solution



Searches related to normal solution preparation pdf filetype:pdf

needed to prepare a specific solution is: Weight in grams = desired molarity x volume needed in litres x GMW Ex: How much weight of NaCl is required to prepare 500mL of 1 M NaCl solution Weight in grams= 1M x 500 mL x 58 45 GMW Weight in grams= 29 225 grams of NaCl to be dissolved in 500 mL of water to make it 1M NaCl solution 2 3 NORMAL SOLUTIONS

How do you prepare a normal solution?

    Normal solutions are prepared by dissolving gram equivalent weight of solute making 1 litre of solution. It means, to prepare 1 liter solution, we have to dissolve the solute equal to the equivalent weight of the solute in grams. Equivalent weight of any chemical is calculated by dividing the molecular weight with its valence.

What is normality of a solution?

    1. Presentation on Normality 2. The strength of solution measured in terms of gram equivalent per litre is called normality. It is denoted by N. A solution having 1 gram equivalent of the dissolved solute in 1 litre of its solution is called normal solution.

How many equivalents are in a normal solution?

    A normal solution contains one equivalent of solute per liter of solution. For acid-base reactions, an equivalent is the amount of a reactant that can produce or consume one mole of hydrogen ions (using the Brønsted-Lowry definition). So, for example, a mole of HCl or NaOH is one equivalent, but a mole of H?SO? or Ca (OH)? is two equivalents.

Preparation of Solutions - Normal solutions

The gramme equivalent of a compound is the number of grammes of the compound which can replace, or is in

any way equivalent to 1 gm of hydrogen.

Normal solution of HCL

Relative molecular mass of HCL is 36.5. In 36.5 g of HCL, there is 1gm of replaceable hydrogen. Therefore the

equivalent of HCL is 36.5.

Normal solution of H2SO4

The relative mass of Sulphuric acid is 98. In 98 gm of H2SO4, there is 2 gm replaceable hydrogen. In 49 gms

of H2SO4, there is 1gm of replaceable hydrogen. Therefore the equivalent of H2SO4 is 49.

Since equivalents of compounds react with one another, the figures obtained above can be used to obtain the

equivalents of bases and salts. This is done by working from the equiation for the reaction between the

compound needed and one whose equivalent is known.

Consider the equation

NaOH + HCL = NaCl +H2O

40g + 36.5g

40 Gm of NaOH is equivalent to 36.5 gm HCL. Therefore the equivalent of NaOH is 40.

Consider the equation

AgNO3 + HCL = AgCl + HNO3

170GM + 36.5 GM

170 gm of AgNO3 is equivalent to 36.5gm HCL. Therefore the equivalent of AgNO3 is 170.

How to prepare a normal solution?

After calculating the equivalent of the compound, the next step is to determine how many gms of the compound

is required for making a given volume of solution of known normality.

4 N NaOH Solution

Required quantity of 4N NaOH 5 Litres (Equivalent is 40) Therefore 1 litre of IN NaOH contains 40gm NaOH. Therefore 1 litre of 4N NaOH contains 4 x 40 gm NaOH = 160 gm NaOH. Therefore 5 litres of 4N NaOH contains 5 x 160gm = 800 gm NaOH.

So 5 Litres of 4N NaOH solution can be prepared by dissolving 800 gms NaOH in 5 litres of distilled water.

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