>> I got 0.104359 mol of H+ <<
I forgot to add the "free" H+ ion that must be present to have a pH of 2.00. The solution must be 0.010 M in free (dissociated) H+ ion to give a pH of 2.00. The moles of free H+ ion is 0.500 L X 0.010 mol/L = 0.0050 mol. Adding this to 0.104359 mol gives 0.109359 mol, or about 0.11 mol H+ to add. Maybe! I need to re-examine this approach in case I am oversimplifying.
Free Forum Hosting
Important Announcement
Prev Message
