posted by Helga .
In the titration of 20.00 mL of 0.175 M NaOH, calculate the number of milliliters of 0.200 M HCl that must be added to reach a pH of 12,55.
I know the answer is 11,9 mL but I don't know how to calculate it.
NaOH + HCl ==> NaCl + H2O
You want pH = 12.55 which means the solution will be quite basic so all of the NaOH has not been neutralized.
14-12.55 = pOH = 1.45 and (OH^-) = -log(OH^-) so (OH^-) = 0.0355M.
If we let x = mL 0.2M HCl added, then mmols OH^- remaining in the titration will be 0.0355*(20.00 + x).
We start with 20.00*0.175 = 3.500 mmols NaOH. As we titrate it the NaOH will be
(3.500-0.2x). Set these equal to each other to obtain
(3.500-0.2x) = 0.0355(20+x) and solve for x = mL 0.2M HCl. Watch the significant figures (I didn't) and check it when you are finished. I worked it out and obtained 11.85 mL which when plugged back into the problem gives a pH of 12.55. I would round that 11.85 to 11.8 but your prof may not round like I do. You may be able to tweak it here and there in the equation but I believe this is the way to work it.