Posted by **Helga** on Friday, November 22, 2013 at 6:04am.

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.

- Chemistry -
**DrBob222**, Friday, November 22, 2013 at 11:57pm
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.

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