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Topic: pH, pKa problem has me stumped!!  (Read 6537 times)

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Offline bolts4life43

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pH, pKa problem has me stumped!!
« on: October 06, 2009, 10:51:37 PM »
Help would be much appreciated! Here's the problem:


An unknown compound, X, is though to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 and 8. When 75 mL of 0.1M NaOH is added to 100 mL of a 0.1M solution of X at pH 2.0, the pH increases to 6.72. Calculate the pKa of the second ionizable group of X.

OK, so I'm not sure the pKa of the carboxyl group is relevant here, since at the initial pH, it will be 50% dissociated. As it goes up to pH 6.72, we will have most of it be dissociated. Thus, I need to find the A-/HA ratio at the new pH to use the Henderson-Hasselbalch equation (pH = pKa + log(A-/HA)). I tried it this way but I'm pretty sure there's something wrong: converted Molar of X solution to moles using the given volume and subtracted the moles added of NaOH (.1M x .075L) to get a pKa of 5.87. Another way I did this that sounds more reasonable is since the compound X is very acidic (initial pH 2) and the NaOH is obviously our base, I set HA = moles of X = (.1M)(.1L) and A- = moles of NaOH = (.1M)(.075L). Using the final pH of 6.72 and plugging into the H-H equation, I got a pKa of 6.84. Again, help would be much appreciated cuz I might be over-thinking this too much!
« Last Edit: October 06, 2009, 11:14:04 PM by bolts4life43 »

Offline Borek

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Re: pH, pKa problem has me stumped!!
« Reply #1 on: October 07, 2009, 04:13:10 AM »
TBH I can't decipher what you did from your description.

As you have correctly stated, initially 50% of the first proton was already neutralized. Then NaOH was added. How much of the added base was used for the full neutralization of the first proton? What excess was left? This excess was used for the neutralization of thse second proton. How far did the neutralization of second proton go? That gives you HA-/A2- ratio, which you just plug into H-H equation.
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