Asked By: Anand sharma
College: Oriental Group of Institutes
Question:
How to prepare 0.5N HCL solution.?
1 Answers
1266
SOHANLAL KUSHWAHA
By adding 41.6 ml HCL in 1000- 41.6 = 958.4 ml of water.. Because.. Normality of a solution is defined as the product of its molarity and n factor i.e. Normality (N)=Molarity × n factor--------(1) (Where n factor is acidity or basicity i.e. the no. of H+ ions or OH- ions released) Here as in HCl we have only one H+ ion So, n factor = 1 Substituting values in (1) 0.5 = M×1
So, M = 0.5
Concentrated HCl is 12 M, which means 12 moles/L or 0.012 moles/mL.
A 0.5 M solution of HCl has 0.5 moles/L. So for 1 liter of solution, you need 0.5 moles of HCl.
To get 0.5 moles of HCl requires 41.67mL conc. HCl:
(0.012 moles/mL)(x mL) = 0.5moles
x = 0.5/0.012
x = 41.67mL
But remember from above:
This is for 1 L of solution, so you need to dilute that 41.67mL of conc. HCL up to a total volume of 1 liter (meaning you add the 41.67mL conc. HCl to 958.33mL water.)
Always add acid to water, not the other way as the dilution of HCl in water produces heat, and you want that heat to be dissipated over as much volume as possi…
Because..
Normality of a solution is defined as the product of its molarity and n factor
i.e. Normality (N)=Molarity × n factor--------(1)
(Where n factor is acidity or basicity i.e. the no. of H+ ions or OH- ions released)
Here as in HCl we have only one H+ ion
So, n factor = 1
Substituting values in (1)
0.5 = M×1
So, M = 0.5
Concentrated HCl is 12 M, which means 12 moles/L or 0.012 moles/mL.
A 0.5 M solution of HCl has 0.5 moles/L. So for 1 liter of solution, you need 0.5 moles of HCl.
To get 0.5 moles of HCl requires 41.67mL conc. HCl:
(0.012 moles/mL)(x mL) = 0.5moles
x = 0.5/0.012
x = 41.67mL
But remember from above:
This is for 1 L of solution, so you need to dilute that 41.67mL of conc. HCL up to a total volume of 1 liter (meaning you add the 41.67mL conc. HCl to 958.33mL water.)
Always add acid to water, not the other way as the dilution of HCl in water produces heat, and you want that heat to be dissipated over as much volume as possi…