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The normality of commercial grade hydrochloric acid (HCl) can vary depending on the concentration specified by the manufacturer. Hydrochloric acid is commonly available in different concentrations, such as 37% or concentrated hydrochloric acid. To determine the normality, it is essential to know the molarity (moles of solute per liter of solution) and the number of equivalents of the acid.

Normality (N) is related to molarity (M) by the equation:

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=

�

×

�

N=n×M

where:

�

N is the normality,

�

n is the number of equivalents,

�

M is the molarity.

For hydrochloric acid (HCl), which is a monoprotic acid (donates one proton), the number of equivalents (

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n) is equal to 1.

Therefore, if you know the molarity of the commercial grade hydrochloric acid, you can determine its normality using the equation mentioned above. It's important to check the product label or contact the manufacturer for the specific concentration of the hydrochloric acid you are using.

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The normality of a HCl commercial solution (hydrochloric acid) is generally 35-38 %.

normality of the hydrochloric acid available is app. 12.4 N, whereas for water it will be1.0 both for its conjugate acid as well as conjugate base.

Q: What is the normality of commercial grade HCl?

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Normaly HCl is supplied 35.5% w/w in 2.5 ltr bottles. Its Specific gravity is 1.18 g/ltr. First we convert weight to volume using provided inputs. Now taking all units in CGS g HCl =2500 X1180 =2950 g HCl Further for 35.5 % (means 35.5 in 100) then g HCl =0.355X2950=1047.25 Mole HCl=1047.25/36.5=28.69 Mole Normality and molarity will be same in case of HCl. Molarity=28.69/2.5=11.476 Approx.11.5

g HCl solution = 2500 mL of HCl * 1 liter/1000 mL * 1190 g/L = 2975 g 37% solution (37 g HCl/100 grams of solution) gives you the grams of HCl: g HCl = 0.37 * 2975 g = 1100.8 g HCl Moles HCl = 1100.8/(36.46 g/mole) = 30.2 moles Therefore the molarity, which equals the normality in this case = 30.2 moles/2.5 L = 12.07 M = 12.07 N If you want to make 100 mL of a 0.1 N solution, Volume of HCl solution needed = (0.1 N * 100 mL) /12.07 N = 0.83 mL Take 0.83 mL of the 37% HCl, and dilute it with water to 100 mL.

1N HCL is the same as 1 Molar HCl. You take the # of H ions and multiply by the molarity to get the Normality. Usually you buy HCl in concentrated form which is 12 Molar or 12 Normal HCL. You need to dilute the concentrated HCl to get the reduced concentration. Use the formula Molarity Initial x Volume Initial = Molarity Final x Volume Final ex. 12 M HCL x 10 ml = 1 M x 120 ml. So take 10 ml of concentrated HCl and add enough water to make 120 ml. This will give you 120 ml of 1 M (which is 1N) HCl. Venkat Reddy

formula for neutralizatrion is volume of acid X normality of acid = volume of base X normality of base so (0.3)(3) should equal (4)(volume) which is .225L. However, Ca(OH)2 contains 2 moles of OH resulting division of total volume needed by 2. Thus, the answer becomes .1125L or 112.5ml.

To find the normality of ferrous ammonium sulfate, use this formula: Normality of Ferrous Ammonium Sulfate = (Volume of Potassium Dicomate, ml) X 0.250N Divided by Volume of Ferrios Ammonium Sulfate, mL

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Normaly HCl is supplied 35.5% w/w in 2.5 ltr bottles. Its Specific gravity is 1.18 g/ltr. First we convert weight to volume using provided inputs. Now taking all units in CGS g HCl =2500 X1180 =2950 g HCl Further for 35.5 % (means 35.5 in 100) then g HCl =0.355X2950=1047.25 Mole HCl=1047.25/36.5=28.69 Mole Normality and molarity will be same in case of HCl. Molarity=28.69/2.5=11.476 Approx.11.5

The normality is 10,8.

Normality = molecular mass/equivalent mass 1 mole (M) HCl = 1M HCl × 36.5/18.25 = 2 Ans

The HCL concentration is 1.2M or 1.2N

NaOH

This can only be established by accurate standard titration with sodium carbonate (p.a.)

0.2 N HCl solution means 0.2 equivalents of HCl dissolved in 1 litre of water. Normality = Molarity x n-factor => Molarity =Normality/n-factor=0.2/1=0.2 M 0.2 moles should be present in 1 litre of solution. 0.2moles =0.2 x 36.5 = 7.3 grams of HCl =>Dissolve 7.3 grams of HCl in 1 litre if water to get a 0.2N solution.

The answer is 15,039 g hydrogen chloride (HCl).

Normality is the same as Molarity assuming there is only 1 exchangeable proton. Hence, 5 Normal = 5 Molar, which is 5 mol of HCl per Liter of solvent (water)

g HCl solution = 2500 mL of HCl * 1 liter/1000 mL * 1190 g/L = 2975 g 37% solution (37 g HCl/100 grams of solution) gives you the grams of HCl: g HCl = 0.37 * 2975 g = 1100.8 g HCl Moles HCl = 1100.8/(36.46 g/mole) = 30.2 moles Therefore the molarity, which equals the normality in this case = 30.2 moles/2.5 L = 12.07 M = 12.07 N If you want to make 100 mL of a 0.1 N solution, Volume of HCl solution needed = (0.1 N * 100 mL) /12.07 N = 0.83 mL Take 0.83 mL of the 37% HCl, and dilute it with water to 100 mL.

Acid solutions are typically made in laboratories from commercially available acids which are supplied with specifications as to their physical and chemical properties including their concentrations. Calculations: For example, if you wish to use concentrated HCl that just arrived in your lab yesterday to make (say) 1N HCl, you will need to know the normality of the available solution. Suppose you know that it is 37% HCl (and all other information is missing); this means 37 mL HCl in 100 mL solution, M.W. of HCl = 36.5, Density = 1.185; 1.185 g HCl occupies 1 mL volume, 37 mL HCl corresponds to 43.84 g HCl. If the commercially available solution is 43.84 g HCl in 100 mL solution and you know that 36.5 g HCl in 1000 mL solution makes 1N (also 1M HCl) solution, then you have a 12N (also 12M) solution in your hands. So that means you must dilute it 12 times to get a 1N HCl solution. Titration: If you don't have a new solution at hand and are not sure about how correctly the HCl reagent was stored over many years, it would be prudent to measure the concentration of acid by titration provided you have fresh (reliable) base solutions at hand and reliable indicators.

normality of acetic acid=%acetic acid*density*1000/mol. wt.*100 = 4*1.05*1000/60*100 =0.7 N