Help needed- How do I do this pH neutralisation ?! Watch

Bien
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#1
Report Thread starter 11 years ago
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Very stuck here.

I have a question which I have no clue what to do .

A mass, m, of sp;od KOH is added to 755cm3 of 0.0120 M HCl. The pH after this addition is 11.60 at 25 C. The volume of the resulting solution is still 755cm3.

1- calculate the number of moles of OH- ions needed to neutralise exactly the H+ ions present in the 755cm3 of 0.0120 HCl.

Firstly I worked out the pH of the HCl which I got to be 1.92.

Then do I do : 7- 1.92? To get 5.08

Then do the kw = [H+][OH-]
and do 5.08 = 1 x 10 -14 (superscript) / OH
OH = 1 x 10-14 / 5.08


I'm lost here- please help.
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charco
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Why not just calculate the molarity of the KOH solution after the addition. The pH after this addition is 11.60 and as pH + pOH = 14
Then pOH = 2.4

Therefore the OH- ion concentration = 10^-2.4
This is also the concentration of the KOH

Now you know the volume of solution = 755cm3 and the molarity of the KOH so you can work out the number of moles.

Now work out the no of moles of HCl at the beginning = 0.755 x 0.012

As KOH + HCl --> products (i.e. a 1:1 reaction) then KOH needed to neutralise this number of moles of HCl is the same

Therefore total no of moles of KOH added = (0.755 x 10^-2.4) + (0.755 x 0.012)

And from the moles you can work out the mass...
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