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 HCl (approx.7.5ml) and dilute to 15ml with deionized water.

Please explain how to perform this procedure. Conc HCl or 1M HCl? I do not understand it.
Thanks.

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