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Topic: Henderson Hasselbalch solubility question  (Read 2291 times)

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Offline CCG88

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Henderson Hasselbalch solubility question
« on: October 10, 2017, 02:19:09 PM »
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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Offline XeLa.

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Re: Henderson Hasselbalch solubility question
« Reply #1 on: October 10, 2017, 06:03:43 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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Offline CCG88

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Re: Henderson Hasselbalch solubility question
« Reply #2 on: October 11, 2017, 08:47:28 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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Offline sjb

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Re: Henderson Hasselbalch solubility question
« Reply #3 on: October 11, 2017, 08:59:43 AM »
If I had the equation x = log 100000, do you know how to calculate x?
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Offline CCG88

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Re: Henderson Hasselbalch solubility question
« Reply #4 on: October 11, 2017, 11:51:16 AM »
Take the log of 100000 which is 5, therefore x=5
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Offline sjb

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Re: Henderson Hasselbalch solubility question
« Reply #5 on: October 11, 2017, 12:14:34 PM »
So, if you have 6 = log x; can you therefore calculate x?
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Offline CCG88

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Re: Henderson Hasselbalch solubility question
« Reply #6 on: October 11, 2017, 12:33:04 PM »
Take the antilog of 6 which is 1000000, therefore x=1000000
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