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1N HCl Solution
A solution is mainly composed of two components; solute and solvent.
For example; in a solution of salt in water, salt is the solute and water is the solvent.
For a solution, in which both the component have same physical state as that of solution, the component which present in excess is called as solvent and other one is known as solute.
The concentration of solution can be expressed in various ways like mole fraction, Molarity, normality, molality and part per million.
Out of these expressions, normality is the most common way to express concentration of solution.
Normality of solution is defined as the number of gramequivalents of the solute present in one liter of solution and represented as N.
The unit of normality is gequ/L.
The solution containing one gramequivalent of solute in one L solution is known as a normal solution.
Mathematically Normality can be written as;
Normality (N) = WB(g) x 1000
EB x V (ml)
Where; WB(g) = mass of solute B in gm
EB = Equivalent mass of solute B
V = volume of solution (ml)
For the preparation of normal solution, we must be familiar with equivalent mass of given substance.
The gram equivalent weight can be calculated by dividing the molecular weight of substance in gm by its valence (n);
Equivalent weight (EW) = Molecular weight (gm) / n
A "one Normal" solution (1 N) contains one “gram equivalent weight” of substance , toppedoff to one liter of solution.
Weight of solute
N = __________________________________________________
Equivalent weight of solute × Volume (in ml) of dilution
For the formation of normal solution from pure or nonaqueous acids like hydrochloric acid (HCl), the equivalent weight of an acid is equals to the molecular weight divided by basicity of acid which is equals to the number of replaceable hydrogen atoms in the reaction.
The basicity of Hydrochloric acid (HCl) is one as it has one replaceable hydrogen ion (H^{+}).
Therefore the equivalent weight of pure HCl, is 36.46 and for the preparation of a 1N solution 36.46 grams of the pure Hydrochloric acid (HCl) must be added to per liter of solvent.
Generally in laboratory concentrated aqueous solution of acids is used for the preparation of desired solution.
For the preparation of standardised Normal solution from aqueous laboratory concentrated reagent, the specific gravity (grams/ml) and the percentage composition by weight of the substance are necessary.
For example; for the preparation of 2 L of a 1 N hydrochloric acid solution from a given solution of HCl which has density of 1.188 g/ml and containing 38% HCl by weight,
we needed;
Molecular weight of HCl = 36.461
Valence (n) of hydrogen = 1
Weight of HCl needed = 1 eq/Liter × 2.0 Liters × 36.461 g/eq
Or 1 eq × 2000 ml × 36.461 grams
Weight of HCl needed = _______ ____________
1000 ml eq
= 72.92 grams
Since the weight of HCl = V(ml) × 1.188 grams × 0.38
Therefore 14.5844 grams = V(ml) × 1.188 grams × 0.38
And V(ml) = 72.92 grams × ml
_________________
1.188 grams × 0.38
V(ml) = 161.53 ml
So 161.53 ml of concentrated HCl has to add to water and make total volume of 2 L for getting 1 N solution of HCl.
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Question:
Answers:nivi=n2v2 n is the normality
Answers:nivi=n2v2 n is the normality
Question:
Answers:It depends on what you have to start with. If you have 1N HCl, then you need to add 9 mL of water to each 1 mL of 1N HCl. This is a 1 in 10 dilution. (1/10) of 1N = 0.1N If you have concentrated HCl, then you need to look on the bottle for the percentage by weight, and the density. From these, the molarity, or normality (which is the same for HCl), can be calculated. Generally speaking, conc. HCl is about 12N, so if you wanted to use that value, then you'll need to dilute 1 in 120, which is the addition of 119 mL of water to each 1 mL of conc. HCl. To calculate the exact value of the normality of the conc. HCl, the following formula will help : Molarity = Normality = 10 * P * D / M where : P = percentage HCl by weight. (%) D = density (g/mL) M = molecular weight ( = 36.46) As an example, if you calculate the normality to be, say, 12.2N, then you need to dilute 1 mL of conc. HCl with 11.2 mL of water to make 1N HCl. From there, another 1 in 10 dilution will get you 0.1N HCL.
Answers:It depends on what you have to start with. If you have 1N HCl, then you need to add 9 mL of water to each 1 mL of 1N HCl. This is a 1 in 10 dilution. (1/10) of 1N = 0.1N If you have concentrated HCl, then you need to look on the bottle for the percentage by weight, and the density. From these, the molarity, or normality (which is the same for HCl), can be calculated. Generally speaking, conc. HCl is about 12N, so if you wanted to use that value, then you'll need to dilute 1 in 120, which is the addition of 119 mL of water to each 1 mL of conc. HCl. To calculate the exact value of the normality of the conc. HCl, the following formula will help : Molarity = Normality = 10 * P * D / M where : P = percentage HCl by weight. (%) D = density (g/mL) M = molecular weight ( = 36.46) As an example, if you calculate the normality to be, say, 12.2N, then you need to dilute 1 mL of conc. HCl with 11.2 mL of water to make 1N HCl. From there, another 1 in 10 dilution will get you 0.1N HCL.
Question:I have commercially available 35% mass/mass HCl stock solution, I want to prepare 1N HCL from it how should i do? I wish to know molar calculations..
Thanks in advance..
Answers:HCl has a molecular mass of 36.5, meaning a 35% solution is roughly 10N. So just dilute it by a factor of ten. If you're making a litre of the 1N solution, use 100ml of the stock and 900ml of water.
Answers:HCl has a molecular mass of 36.5, meaning a 35% solution is roughly 10N. So just dilute it by a factor of ten. If you're making a litre of the 1N solution, use 100ml of the stock and 900ml of water.
Question:
Answers:pH 1.5 has a [H+] = 0.0316 M For HCl 1N = 1M Liters x Molarity = Moles 31.6 mL diluted to 1 liter will yield a 0.0316 M HCl solution
Answers:pH 1.5 has a [H+] = 0.0316 M For HCl 1N = 1M Liters x Molarity = Moles 31.6 mL diluted to 1 liter will yield a 0.0316 M HCl solution