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Describe how you would prepare 2.00 L of an aqueous solution having pH = 3.60, if you had a supply of 0.100 M HCl available.
Given that pH = log[H3O+], the desired concentration of hydronium ion is
[H3O+] = 10pH = 103.60 = 2.5×104 M.
Since HCl is a strong acid, i.e., HCl(aq) + H2O(l) H3O+(aq) + Cl(aq),
[H3O+] = [HCl] = 0.100 M in the given solution, so this needs to be diluted to obtain the desired concentration.
Recalling Mconc × Vconc = Mdil × Vdil, the amount of 0.100 M HCl required is
Mconc = 0.100 M
Mdil = 2.5×104 M
Vdil = 2.00 L
so Vconc = (2.5×104 M × 2.00 L)/(0.100 M) = 0.0050 L = 5.0 mL.
Thus to obtain the pH = 3.60 solution, add 5.0 mL of the 0.100 M HCl solution to a 2.00 L volumetric flask and add water to the mark.