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Topic: pH of Solutions  (Read 27439 times)

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Offline fledarmus

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Re: pH of Solutions
« Reply #30 on: October 04, 2011, 08:07:18 AM »
Well, there's a little more to it than that  ;)

How much water? And what would be left over?

This is a balanced equation/limiting reagent problem now
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Offline specialk08

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Re: pH of Solutions
« Reply #31 on: October 04, 2011, 12:07:26 PM »
Honestly, you've lost me haha.
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Offline Borek

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Re: pH of Solutions
« Reply #32 on: October 04, 2011, 12:15:09 PM »
Quote from: fledarmus on October 04, 2011, 08:07:18 AM
This is a balanced equation/limiting reagent problem now

No, this is salt, so they are in equimolar quantities.
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Offline specialk08

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Re: pH of Solutions
« Reply #33 on: October 04, 2011, 12:27:49 PM »
Care to elaborate for those of us not able to do this problem :p
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Offline fledarmus

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Re: pH of Solutions
« Reply #34 on: October 04, 2011, 01:08:57 PM »
Oops, sorry, I completely misread your previous post - I thought the question had changed. Never mind...  :-[
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Offline specialk08

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Re: pH of Solutions
« Reply #35 on: October 04, 2011, 05:53:02 PM »
So is anyone able to point me in the proper direction?  :)
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Offline Dan

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Re: pH of Solutions
« Reply #36 on: October 04, 2011, 06:13:19 PM »
OK, so you know that:

H+ + -OH --> H2O

...and you know [H+] generated by 0.1 M anilinium, and you know [-OH] generated by 0.1 M fluoride. What is left  over?
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Offline specialk08

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Re: pH of Solutions
« Reply #37 on: October 04, 2011, 06:32:44 PM »
Ummmmm just water? Why is this confusing me so much haha :/
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Offline Dan

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Re: pH of Solutions
« Reply #38 on: October 05, 2011, 03:20:58 AM »
How much water? is the ratio of H+ to -OH 1:1 or is one of them in excess over the other. What is left behind after the water forms?
« Last Edit: October 05, 2011, 10:42:37 AM by Dan »
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Offline Borek

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Re: pH of Solutions
« Reply #39 on: October 05, 2011, 08:20:03 AM »
Dan, don't you forget that after neutralization both substances will dissociate further to reach a new equilibrium?
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Offline Dan

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Re: pH of Solutions
« Reply #40 on: October 05, 2011, 10:43:13 AM »
Yeah, this is the part I'm not too sure about. Perhaps you can take over Borek?
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Offline Borek

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Re: pH of Solutions
« Reply #41 on: October 05, 2011, 05:26:48 PM »
Geez, I know the answer, but I don't see how to get it.

In general it seems like pH = 0.5*(pKa1+pKa2), same form like in the case of amphiprotic salt, although meaning of pKa1 and pKa2 is different. But I got stuck deriving this formula. I will try tomorrow again (I hope I won't forget).
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Offline specialk08

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Re: pH of Solutions
« Reply #42 on: October 05, 2011, 06:51:10 PM »
My professor told us to disregard that question, as she was having too many questions about it and realized the text didn't thoroughly cover it.. thanks guys :)
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Offline Borek

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Re: pH of Solutions
« Reply #43 on: October 06, 2011, 07:44:10 AM »
OK, I got it. Actually I am an idiot, as after solving it today I realized I had it solved long ago and the answer was already in my lectures  :-\

BUT - and that's a good news - I never liked the way I derived the formula in the previous version, and today I found a much better way of getting identical result.

http://www.chembuddy.com/?left=pH-calculation&right=pH-salt-simplified
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Offline specialk08

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Re: pH of Solutions
« Reply #44 on: October 06, 2011, 09:19:47 PM »
Thanks for that! I'll keep that on record incase I encounter this again :)
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