You have now specified 12 M HCl, that is not dilute.
The white ppt was NaCl not in solution. You added a lot of Cl ions with the 12M acid, you already had Na ions. So many, they could not go into solution.
The following is part of a procedure for a limit test for sulphates in an NaOH sample:-
Dissolve 3.0g of NaOH in 6ml of deionized water, adjust to pH 7 with 12M HCl (about 6.2ml) and dilute to 15ml with deionized water.
When about 6ml of 12M HCl was added, a white ppt was obtained.
What happened?
Thanks.
Please explain how to perform this procedure. Conc HCl or 1M HCl? I do not understand it.
Thanks.
Chemistry - urgent - bobpursley, Tuesday, September 21, 2010 at 10:09am
What form of NaOH did you use? Assume granular.
moles NaOH=3/40
moles HCl needed= 3/40=volume*Molarity
so lets see what 7 ml will neutralize
molarity=3/40/.007= or about 10Molar. 32Percent HCl is about 10M
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