Ferric sulphate

I'm trying to make a concentrated solution (about over 40% )of ferric
sulphate from FeSO4, H2O2 and H2SO4. The result is acceptable except
the pH. It is lower than zero. I don't know why because H+ from H2SO4
will take part in the reaction to form H2O. Is this caused by the
hydrolysis of Fe3+ ? How could I make the result better ? Should I add
some other chemicals to increase the pH while I need the pure form of
ferric sulphate ? Best regards.
TMT

>From: [Only registered users see links. ] (Thieu Minh Triet)



The pH of 40 % solutions is a meaningless response of the pH probe to the
conditions present in the solution ...In fact, if you changed probe
types(Ag/AgCl, calomel,etc.) ... you'd get different answers .. To understand
more read up on physical properties of concentrated solutions ...The
concentrations and activities are nowhere near equal ... other effects like
viscosities etc. also come into effect for probe response ...
To measure how much free acid is present, generally one samples ... dilutes
1-50 or so and titrates either electrochemically to the end point or if
possible using an indicator that will change before you start hydrolyzing the
metal salt ...



Be seeing you
In the Village
Number 6

[Only registered users see links. ] (Thieu Minh Triet) wrote in message news:<71e1c728.0312310222.c5e07f5@posting.google.c om>...

Ferric sulfate is generally available as a technical grade. It is not
supplied as an analytical grade. Hence there is no information
available as to the typical free acid present. If you want ot
determine the free H2SO4 present, here is the procedure from "AnalaR
Standards for Laboratory Chemicals" BDH 8th edition, based on the
procedure for free acidity in ferric chloride.

Dissolve 6g of KF in 25mL of water in a platinum or polyethylene
vessel, add 0.1 ml of phenolphthalein solution and titrate with 0.1N
NaOH to a red color. To this solution add a solution of 2g of the
sample in 15ml of water, dilute to 50 ml with water and set aside for
3 hours. Filter through a dry paper and titrate 25ml of filtrate with
0.1N NaOH to the same red color.
1 ml of 0.1N NaOH is equivalent to 4.904mg H2SO4.