You will need to know the molarity of the solution you work from. Remember when working with acid - always put acid to the water, never vice versa.
If you have a 1M solution you will need to put 20 times the amount of water.
If you have a 5M solution you will need to 100 times the amount of water.
It is more difficult if the basic solution is given in percentage. In this case you have to calculate how much HCl the solution contains.
Eks::
A 37% HCl has a density of 1,19g/cm3 and a molarity of 12,1 (12,1M). so the amount of water will be 242 times more to get the specified solution.
Be very careful if you are working with this stuff - you could get injuries on you skin, eyes and lounges.
Weigh accurately 0.5gm of Anhydrous Sodium Carbonate and dissolv in 20 ml of distilled water.
Titrate this solution with the prepared 0.5N Hydrochloric Acid using methyl orange as an indicator until the appearance of reddish-yellow color as end point. Boil for 2 minutes and continue the titration until the appearance of reddish-yellow color.
1ml of 0.5N Hydrochloric Acid is equivalent to 26.5 mg of anhydrous Sodium Carbonate.
CALCULATIONS:
Wt. of Anhydrous Sodium Carbonate in gm x 1000 x 0.5
A) Normality = ---------------------------------------------------------------------26.5 x Volume of 0.5N HCl consumed
0.05 M means 0.05 mol of HCl in 1 Litter of solution
Add 200 mls of the 1.0N HCl solution to a 1 liter volumetric flask. Make up to the mark with water. Standardize against a known weight of Sodium Carbonate.
HCl is a gas. It's therefore very difficult to accurately weigh out the proper amount to use to make a solution of any given concentration, and as the solution sits, some HCl may escape as gas. In contrast, it's very easy to weigh out a solid base to high precision, and use a solution made from it to standardize the only-approximately-known HCl concentration.
About 13M. You can assume it is 13M if you don't need an exact concentration (like if you need a ~1M HCl solution for an extraction or whatever) but if you need an exact concentration (for a titration, for example) then you will need to standardize your HCl first.
If the solution is not a buffer, the HCl will react with the solution to form a product.
1N HCl is also 1M HCl because it is mono-protic. Therefore 36.5 g of HCl is required per liter or 3.65%. Simply take 100 g of 37% HCl and make up to the 1 liter mark on the volumetric flask. Check the value by titration against 1M NaOH. It should be perfect. If very slightly strong dilute very slightly (calculate) with water and re-standardize.
HCl is a gas. It's therefore very difficult to accurately weigh out the proper amount to use to make a solution of any given concentration, and as the solution sits, some HCl may escape as gas. In contrast, it's very easy to weigh out a solid base to high precision, and use a solution made from it to standardize the only-approximately-known HCl concentration.
Add 200 mls of the 1.0N HCl solution to a 1 liter volumetric flask. Make up to the mark with water. Standardize against a known weight of Sodium Carbonate.
HCl is a gas. It's therefore very difficult to accurately weigh out the proper amount to use to make a solution of any given concentration, and as the solution sits, some HCl may escape as gas. In contrast, it's very easy to weigh out a solid base to high precision, and use a solution made from it to standardize the only-approximately-known HCl concentration.
About 13M. You can assume it is 13M if you don't need an exact concentration (like if you need a ~1M HCl solution for an extraction or whatever) but if you need an exact concentration (for a titration, for example) then you will need to standardize your HCl first.
If the solution is not a buffer, the HCl will react with the solution to form a product.
1N HCl is also 1M HCl because it is mono-protic. Therefore 36.5 g of HCl is required per liter or 3.65%. Simply take 100 g of 37% HCl and make up to the 1 liter mark on the volumetric flask. Check the value by titration against 1M NaOH. It should be perfect. If very slightly strong dilute very slightly (calculate) with water and re-standardize.
(7 mL)(X M HCl) = (27.6 mL)(0.170 M NaOH)7X = 4.692X = 0.7 M HCl==========
If it is only HCL, yes.
HCl is ionozed in aq solution HCl + H2O = H3O(+ CHARGE) + Cl (- CHARGE) HCl FORMULA WILL REAMAIN HCl OT WILL BOT CHANGE
Molarity = moles of solute/liters of solution Molarity = 0.597 moles HCl/0.169 liters = 3.53 M HCl ------------------
Yes. HCl is a strong electrolyte.
In solution with a pH of 1 [H+] is 0.1M. Since HCl is a strong acid [HCl] will also be 0.1M. So, in 1 liter of solution you will have 0.1 mol of HCl.