If I got 37% concentration of HCl how can I have 5% concentration of it?
In order to reduce the percentage of a chemical in a solution, first you need to use the general dilution equation which is:
(C1)(V1)=(C2)(V2)
Whereby C1 and C2 are concentration of the chemical before and after the dilution (reduction of concentration), respectively. In your case, C1 is 37% and C2 is 5%. Assume that you have 100 ml of 37% HCl in water, and you want to dilute these 100 ml of HCl solution using water as diluent to achieve a solution at lower HCl concentration of 5%, the equation can be re-written as:
(37%)(100 ml) = (5%)(V2)
Solving for V2, the final volume (V2) = 740 ml
The final volume includes the initial volume, in other words : V2 = V1 + V(additional water)
In your case: 740 ml = 100 ml + V(additional diluent)
Thus, V(additional diluent) = 640 ml of water required
To put long story short, add 640 ml of water to 100 ml of a solution of 37% HCl in water and mix them well, to achieve a final solution with volume of 740 ml whereby the HCl is 5% only.
You do below calculation for example 100ml of 5% HCl
5%×100 ml=37% × V ml then V ml = 5×100/37= 13.53 ml
In order to reduce the percentage of a chemical in a solution, first you need to use the general dilution equation which is:
(C1)(V1)=(C2)(V2)
Whereby C1 and C2 are concentration of the chemical before and after the dilution (reduction of concentration), respectively. In your case, C1 is 37% and C2 is 5%. Assume that you have 100 ml of 37% HCl in water, and you want to dilute these 100 ml of HCl solution using water as diluent to achieve a solution at lower HCl concentration of 5%, the equation can be re-written as:
(37%)(100 ml) = (5%)(V2)
Solving for V2, the final volume (V2) = 740 ml
The final volume includes the initial volume, in other words : V2 = V1 + V(additional water)
In your case: 740 ml = 100 ml + V(additional diluent)
Thus, V(additional diluent) = 640 ml of water required
To put long story short, add 640 ml of water to 100 ml of a solution of 37% HCl in water and mix them well, to achieve a final solution with volume of 740 ml whereby the HCl is 5% only.
I would do similar to what Ajay explained.
so if you need 1% of 35% you need to:
divide (1X100)/ 35= 2.85 ml
This is the volume you have from 35%.
Then:P
100-2.85=97.15 mls would be the volume you need to add to your 35% to give you 100mls of 1%.
Is that clear?
Simin.
C1 x V1= C2 x V2
C1 is the initial concentration(37%) C2 is the diluted concentration(5%)
for example if you want to prepare 100ml of 5% solution you do the following:
37 x V1= 5 x 100 => V1=13.51ml
So you must take 13.5 ml out of your initial solution (37%) and add them to a 100 ml volumetric flask then add distilled water until the bottom of the meniscus is level with the horizontal line on the neck of the flask.
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