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Topic: HCl NaOH Titration  (Read 2011 times)

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Offline vanderwaals

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HCl NaOH Titration
« on: April 05, 2017, 09:38:51 PM »
25.0 mL of 0.10 M HCl solution is titrated with 0.10 M of NaOH solution. The final pH of the solution
is 12.00. How many mL of NaOH solution are added?
A. 25.0 B. 27.8 C. 55.6 D. 50.0

The answer is (C) 55.6 mL, but I keep getting 30.6 mL when I do the problem for some reason. I calculated that the concentration of OH- has to be 0.01 M for the pH to be 12.00, so ((the volume of NaOH x 0.10 M) - (total moles of HCl)) / (the total volume) must equal 0.01 M, and this only seems to work when 30.6 mL of NaOH is added. What am I doing wrong?

Offline XeLa.

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Re: HCl NaOH Titration
« Reply #1 on: April 06, 2017, 01:56:11 AM »
I get the same answer. I don't understand their logic. The final volume of the solution should be 55.6 mL, however, the volume of added 0.1 M NaOH solution should be 30.6 mL. Assuming, of course, that the solutions are additive.
« Last Edit: April 06, 2017, 02:28:13 AM by XeLa. »

Offline Borek

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Re: HCl NaOH Titration
« Reply #2 on: April 06, 2017, 03:37:48 AM »
Obvious mistake, 30.6 mL is a correct answer.

After adding 55.6 mL of the NaOH solution pH is 12.54
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Offline vanderwaals

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Re: HCl NaOH Titration
« Reply #3 on: April 06, 2017, 08:52:38 PM »
Cool thanks!!

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