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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: CCG88 on October 10, 2017, 02:19:09 PM

Title: Henderson Hasselbalch solubility question
Post by: CCG88 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.

Title: Re: Henderson Hasselbalch solubility question
Post by: XeLa. 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

Title: Re: Henderson Hasselbalch solubility question
Post by: CCG88 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

Title: Re: Henderson Hasselbalch solubility question
Post by: sjb on October 11, 2017, 08:59:43 AM

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

Title: Re: Henderson Hasselbalch solubility question
Post by: CCG88 on October 11, 2017, 11:51:16 AM

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

Title: Re: Henderson Hasselbalch solubility question
Post by: sjb on October 11, 2017, 12:14:34 PM

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

Title: Re: Henderson Hasselbalch solubility question
Post by: CCG88 on October 11, 2017, 12:33:04 PM

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