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Author Topic: Henderson Hasselbalch solubility question  (Read 549 times)

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CCG88

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Henderson Hasselbalch solubility question
« on: October 10, 2017, 08:19:09 AM »

Hello everyone,

I am having trouble understanding the problem in the attached photo. Does anyone know how the answer was actually calculated ? They have simply stated the answer but no explanation as to how they arrived at this. Any help would be greatly appreciated.
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XeLa.

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Re: Henderson Hasselbalch solubility question
« Reply #1 on: October 10, 2017, 12:03:43 PM »

Well, what do you know about the quotient of the Henderson-Hasselbalch equation (the logQ part)? You know how many ions are dissociated (the solubility) at a pH of 8.0, what else forms the quotient?

XeLa
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CCG88

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Re: Henderson Hasselbalch solubility question
« Reply #2 on: October 11, 2017, 02:47:28 AM »

Thanks for the reply. Honestly, I have attempted to work it out but I have no idea, sorry. This stuff totally confuses me
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sjb

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Re: Henderson Hasselbalch solubility question
« Reply #3 on: October 11, 2017, 02:59:43 AM »

If I had the equation x = log 100000, do you know how to calculate x?
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CCG88

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Re: Henderson Hasselbalch solubility question
« Reply #4 on: October 11, 2017, 05:51:16 AM »

Take the log of 100000 which is 5, therefore x=5
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sjb

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Re: Henderson Hasselbalch solubility question
« Reply #5 on: October 11, 2017, 06:14:34 AM »

So, if you have 6 = log x; can you therefore calculate x?
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CCG88

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Re: Henderson Hasselbalch solubility question
« Reply #6 on: October 11, 2017, 06:33:04 AM »

Take the antilog of 6 which is 1000000, therefore x=1000000
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