I'm stucked with a problem of analytics since two days ago.
I've never seen a problem like this and I've tried to solve it with many formulas and reasonable thinking but it always fail apparently.
I need to prepare a 2.5 N HCl from two already prepared solutions which are 3N HCl and 0.5N HCl.
I gave it a shot with the NiVi=NfVf but since it's an unusual dilution it didn't work.
So i figured if i assume that the half of the water in the 0.5 N HCl solutions is available I just have to apply the formula NiVi=NfVf this way:
Nf: 2.5 N
Vf: 1 L
And the result was 833.33 mL so my approach was: If I could use pure water I'd have to use 166.67 mL. But in this case water is forbidden, so I'll use the 0.5 N HCl solution as the diluent, but it only have 0.5 of H2O molecules usable. So I reduce the 833.33 mL 3N HCl solution to half and decided to dilute to the flask mark with 583.4 mL of 0.5 N HCl. But of course it didn't work.
My first questions is: Is it possible to prepare this kind of solutions?
Yes: what kind of special or modifications formula works.
Thanks for your reading and response.