So I have a 1 liter solution with pH = 7,5 and I want to decrease the pH to 3,9. I have a 2M HCl solution to do this. How much HCl solution do I need to add?
My calculation:
pH is the negative log of the hydrogen ion concentration in mol/L
pH = -log([H+]) --> [H+] = 10^-(pH)
In the beginning we have pH = 7,5, so the [H+] concentration is 10^-7,5 = 3,16*10^-8 mol/L
The goal is pH = 3,9, so the [H+] concentration is 10^-3,9 = 1,26*10^-4 mol/L
Required [H+] concentration to decrease pH = 1,26*10^-4 - 3,16*10^-8 = 1,26*10^-4 mol/L
HCl --> H+ + Cl-, so since the stoichiometric are 1:1, the required HCl concentration is also 1,26*10^-4 mol/L
I have a 2 mol/L solution, so I need 1,26*10^-4 / 2 = 6,29*10^-5 liter of this 2M solution to decrease the pH from 7,5 to 3,9 of the original solution.
Is this correct?