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Author Topic: Calculating pH after the equivalence point  (Read 10238 times)

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UG

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Calculating pH after the equivalence point
« on: August 20, 2009, 11:04:11 PM »

I am slightly confused about this matter.
I was reading this table:


After the equivalence point, the last figure: Base added = 20 mL [OH-] = 20/30
This is where I am confused. So at this point 20 mL of base has been added right? And 10mL has already reacted with H3O+ to form H2O right? Therefore, does that not mean that only 10mL of base is in excess? So shouldn't the hydroxide concentration be 10/30?
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UG

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Re: Calculating pH after the equivalence point
« Reply #1 on: August 20, 2009, 11:16:07 PM »

Is it just a typo maybe?  :-X
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Borek

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Re: Calculating pH after the equivalence point
« Reply #2 on: August 21, 2009, 01:14:25 AM »

0.5/20.1 is wrong as well. Typos IMHO.
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UG

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Re: Calculating pH after the equivalence point
« Reply #3 on: August 21, 2009, 01:52:36 PM »

Thanks Borek,
Just to clear something up, if I am titrating a weak acid with a strong base, lets say [HA]=0.15 mol L-1 and [NaOH]=0.15 mol L-1. At the equivalence point, 10mL of NaOH has been added to 10mL of HA. Here is my question I want to clear up: after the equivalence point, when 20mL of NaOH has been added altogether, how would you calculate the pH?
This is what I did; I assumed that all of the OH- ions had been neutralised so therefore the amount of NaOH in excess is 10mL, since the total volume of the solution has been tripled, the concentration has been diluted 3 fold, so [NaOH] is now 0.05 mol L-1, is this correct? The pH would be 12.7.
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Borek

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Re: Calculating pH after the equivalence point
« Reply #5 on: August 21, 2009, 10:11:05 PM »

That was basically what the first link said  :)

"After equivalence point titration curve shape is calculated from the excess of the titrant added - thus it will be just pH of excess titrant in the case of acid-base titration."
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