Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: jbays1973 on October 17, 2004, 03:19:06 PM
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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.
jim
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38 % HCL shows density 1.19 g/mL. Hence 1000 mL weights 1190 grams and contains 0.38 x 1190 = 452.2 g of HCl (12.39 moles per 1L).
Using equation M1V1 = M2V2 you can obtain
V1=0.1 x 1000 /12.39 = 8,1 mL (for 36 % HCl V1= 8.6 mL)