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### Topic: Finding pH at equivalence point  (Read 3395 times)

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#### wizrak

• Regular Member
•   • Posts: 13
• Mole Snacks: +1/-1 ##### Finding pH at equivalence point
« on: October 22, 2011, 05:52:17 AM »
Titration is carried out in which 0.100 mol L–1 NaOH(aq) is added to 10.0 mL of
0.200 mol L–1 hypobromous acid, HOBr, a weak acid with pKa (HOBr) = 8.69.

Calculate the initial pH of the HOBr solution and the pH at the equivalence point.

I used the fact that n(OBr-) = n(OH-)
and used (Kb x [OBr-] )^1/2 = [OH-] and found the pH.

In the mark scheme it used the formula
[H+] = (Kw xKa/[HOBr])^1/2
and used the fact that [HOBr] = [OBr-]
I do not understand this.
At the equivalence point shouldn't the concentration of [HOBr] = 0 as all of them has been converted to [OBr-]

*ps I perfectly know that my method is a correct method, but what I dont get is that
the fact that the mark scheme used [OBr-] = [HOBr] at the equivalence point

#### Borek ##### Re: Finding pH at equivalence point
« Reply #1 on: October 22, 2011, 04:10:31 PM »
I used the fact that n(OBr-) = n(OH-)

Please elaborate - do you mean number of moles of OBr- produced in neutralization equals number of moles of base added, or something else?

At the equivalence point shouldn't the concentration of [HOBr] = 0 as all of them has been converted to [OBr-]

No - that's not the definition of equivalence point. At equivalence point amount of base added equals amount of acid, which means technically you have just a solution of salt. It doesn't mean concentration of HOBr is zero, as OBr- hydrolyzes. Concentration of HOBr is close to zero, but not equal to.

Show what you did step by step.
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#### wizrak

• Regular Member
•   • Posts: 13
• Mole Snacks: +1/-1 ##### Re: Finding pH at equivalence point
« Reply #2 on: October 22, 2011, 06:48:27 PM »
Quote
Please elaborate - do you mean number of moles of OBr- produced in neutralization equals number of moles of base added, or something else?

Yes that is what i meant.

Step by step of what I did

equiv
n(HOBr) = n(OH-) = n(OBr-)
=0.2 x 10/1000 = 0.1Vof NaOH
Hence Vof NaOH = 20mL
Hence total Volume = 20 + 10 = 30mL
[OBr-] = 0.2 x 10/1000 x 1000/30 = 0.066666molL-1
Kb = Kw/Ka = 4.90 x 10-6
[OH-] = root of Kb x [OBr-] = 5.71 x 10-4
Hence pOH = 3.24
Therefore pH = 10.8

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The school told me to use the forumula of the markscheme but what it really doesnt make sense is  that they used the fact that [OBr-] = [HOBr].
Quote
It doesn't mean concentration of HOBr is zero, as OBr- hydrolyzes. Concentration of HOBr is close to zero, but not equal to.
I do not understand this, as we are assuming that HOBr is fromed by ALL THE OBr- getting protonated by water, as [OBr-] and [HOBr] is the same

#### Borek ##### Re: Finding pH at equivalence point
« Reply #3 on: October 23, 2011, 05:42:09 AM »
Your formula and the marking scheme formula are identical (even if they don't look this way).

I think [HOBr] = [OBr-] is supposed to mean concentration of OBr- at equivalence point should equal initial concentration of HOBr (after taking dilution into account). Lousy notation, but not much more lousy than the one you used (n(OBr-) = n(OH-)) - and the result is equivalent. Both the school approach and your approach mix equilibrium and introduced concentrations/numbers of moles making a mess of the problem.

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