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In my lab guide I read: "Use a $\pu{0.1 N}$ solution of methanol-chloridic acid $(9:1)$ for the anthocyanin extraction", so I tried to calculate the volume of $\ce{CH3OH}$ and $\ce{HCl}$ to prepare it, but I don't know how to do it. I mean, if I push one mole of $\ce{CH3OH}$ and one mole of $\ce{HCl}$ in $\pu{1 liter}$ of water, do I make a $\pu{2 molar}$ solution?

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  • $\begingroup$ What about 9 volumes of CH3OH + 1 volume of 1 M HCl, giving 0.1 M HCl in ( approximately ) 90% (v/v) MeOH ? BTW, I have not ever seen the term chloridic acid. $\endgroup$
    – Poutnik
    Nov 18, 2019 at 3:11
  • $\begingroup$ You mean, a 1 M solution of CH3OH? $\endgroup$ Nov 18, 2019 at 5:05
  • $\begingroup$ Why, I have said 1 M HCl. $\endgroup$
    – Poutnik
    Nov 18, 2019 at 5:06
  • $\begingroup$ So, can I use 2 or 3 M solution of CH3OH? The solution ever going to be 0.1 . Also, I think this is a very good answer. $\endgroup$ Nov 18, 2019 at 5:18
  • $\begingroup$ I doubt they consider CH3OH molarity. In my understanding, the target is to have 0.1 M HCl in 90% CH3OH.(approx.) $\endgroup$
    – Poutnik
    Nov 18, 2019 at 5:20

1 Answer 1

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Of course there are such solutions. Take, as an example, Sigma-Alrich's offer of methanolic solution of HCl (3 mol/L, e.g. here), or the dry ones in diethyl ether (e.g., 2 mol/L here), in cyclopentyl methyl ether (e.g., here), or in 1,4-dioxane (e.g. 4 mol/L, here).

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