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Topic: Molecular Weight and pKa of unknown acid  (Read 10275 times)

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Offline GoldShadow

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Molecular Weight and pKa of unknown acid
« on: April 08, 2007, 11:39:06 PM »
This is the problem I'm given:
When titrating 4.25g of a monoprotic weak acid with 0.350M NaOH the endpoint is found to be 48.0 mL.  The pH after 16 mL of the base solution was added was 5.26.  What is the molecular weight and pKa of the unknown acid?

Determining the molecular weight was easy enough.  First I found the moles of base at the endpoint:
0.350M= (mols of base)/0.048L, and there turned out to be 0.0168 mol OH-.  Since at this point moles of base and acid are equal and since it is a monoprotic acid, there are 0.0168 moles of acid as well.  Molecular weight is thus:
(4.25g)/(0.0168 mol) = 253.0 g/mol

Finding the pKa is what I'm stuck on though.  First I determined the moles of OH- after 16 mL have been added:
0.350M=(mols of base)/0.016L.  Moles of base = 0.00560 mol.  Since originally there were .0168 mol of acid present, the amount left at this point (.0168 - .00560) is .0112 mol.  But since I don't know the original volume of acid, I can't turn this into a concentration.

Also since the pH at this point is 5.26, I know, using pH=-log[H+] that [H+] is 5.5*10-6.

So determining pKa would require me to first figure out Ka, and Ka for HA --> H++A- is:  Ka=[H+][A-]/[HA].  Since the moles of H+=moles of A-, it simplifies to Ka=[H+]2/[HA].

The problem is I know the moles of HA but not the total volume of solution at this point so I can't figure out Ka.  Any help on how to solve this would be appreciated.

Offline Yggdrasil

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Re: Molecular Weight and pKa of unknown acid
« Reply #1 on: April 09, 2007, 01:22:22 AM »
Try considering using the Henderson-Hasselbalch equation to figure out the pKa.  Since it depends on the ratio [HA]/[A- you won't need to figure out the volume (since it will cancel out in the numerator and denominator).

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