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Topic: pH exercise  (Read 4763 times)

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Offline uncreative

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pH exercise
« on: December 21, 2016, 11:28:48 AM »
Calculate the pH that you obtain mixing same volumes of 2 solutions of the same base(Kb = 1.85 * 10^5), one having pH=8.65 and the other one having pH=9.25.

Well, I don't understand the problem. I mean, I know the formula for pH=-log[H+]

and that of Kb, but how can I find the final pH?


Offline Borek

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Re: pH exercise
« Reply #1 on: December 21, 2016, 12:01:55 PM »
Would knowing the base concentration help?
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Offline uncreative

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Re: pH exercise
« Reply #2 on: December 21, 2016, 12:07:21 PM »
Kb= ([B+][OH-])/[BOH]   right?


I can find [OH-] using the pH but then I don't know 2 things in that equation. Where am I wrong?

Offline AWK

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Re: pH exercise
« Reply #3 on: December 21, 2016, 01:22:54 PM »
From pH of weak base with known Ka or Kb (pKa or pKb) you can calculate easily concentration this base in both solutions. after mixing you can find a new pH.

http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base
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Offline uncreative

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Re: pH exercise
« Reply #4 on: December 21, 2016, 01:38:54 PM »
Ok that's true i can simplify the equation,, but still how do I know that [OH-]=[B+]?

Offline AWK

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Re: pH exercise
« Reply #5 on: December 21, 2016, 01:59:22 PM »
Quote
but still how do I know that [OH-]=[B+]?
From reaction of dissociation (neglecting autodissociation of water)
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Offline uncreative

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Re: pH exercise
« Reply #6 on: December 21, 2016, 02:31:01 PM »
so, after i  find [BOH] for both solutions, what is my next step? I mean, I know:

1 [OH-] for solution 1
2 [OH-] for solution 2
3 [BOH] for solution 1
4 [BOH] for solution 2


I can't properly figure out how I find the new pH.

I need the final [OH-], right?

Offline AWK

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Re: pH exercise
« Reply #7 on: December 21, 2016, 02:35:34 PM »
Two solution are mixed (equal volumes)
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Offline uncreative

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Re: pH exercise
« Reply #8 on: December 21, 2016, 02:40:24 PM »
Yes, but the volume is unknown, I need the molar concentration of BOH and if I knew the number of moles i could find the volume, but being it unknown, how do I find final [OH-]?

Offline AWK

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Re: pH exercise
« Reply #9 on: December 21, 2016, 02:48:37 PM »
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Calculate the pH that you obtain mixing same volumes of 2 solutions

Volumes are known! You may take any two the same volumes (eg 10+10, 100+100, 500+500). Moreover, there is shortcut - the same result is obtained from arithmetic mean of both concentration (neglecting volumes) for this data.
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Offline uncreative

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Re: pH exercise
« Reply #10 on: December 21, 2016, 02:57:55 PM »
I'm so sorry, but:

final [OH] = total moles (OH)  /  2V;


solution 1: moles (OH)= [OH]*V;
solution 2: moles (OH)= [OH]*V;


can you tell me why I need to use [BOH]? This is what I don't understand.


P.S.

By "I'm sorry", I mean sorry I keep disturbing you

Offline AWK

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Re: pH exercise
« Reply #11 on: December 21, 2016, 03:06:41 PM »
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final [OH] = total moles (OH)  /  2V;
This is not true. As number [OH-]is not equal to [NH3]
You mix solutions of NH3 obtaining a new concentration of ammonia. And concentration of hydroxide ions depends on concentration of ammonia.
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Offline uncreative

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Re: pH exercise
« Reply #12 on: December 21, 2016, 03:09:59 PM »
[quote And concentration of hydroxide ions depends on concentration of ammonia.

what is the dependance between these 2?

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Offline uncreative

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Re: pH exercise
« Reply #14 on: December 21, 2016, 03:29:22 PM »
But isn't true that

final n moles (OH) = n moles (OH from sol 1) + n moles (OH from sol 2)?

or is this also not true?

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