I have a question regarding the preparation of this solution. We are using 36 - 38 % HCl as our stock for the dilution. I understand that 8.3 mL are required to make the 1 L required. What I do not understand is how this quantity is arrived at. I mean, how was it determined that one needs to add 8.3 mL to a liter of DI water to make 0.1 N HCl? I would appreciate the assistance with this question.