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Topic: Finding Ka1 and Ka2 for Malonic Acid  (Read 25358 times)

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Re:Finding Ka1 and Ka2 for Malonic Acid
« Reply #15 on: May 06, 2005, 10:24:16 PM »
i made a substraction error to find the protonic hydrogen available.

protonic hydrogen available
= 2 x intial moles of malonic acid - moles of NaOH added
= 2 x 19.5/104 - 12.5/40
= 0.0625moles

protonic hydrogen available per L of solution
= 0.0625/0.250
= 0.25 moles/L

after i corrected my substraction error, i realise i arrive at the same answer as Borek. You may make reference to the corrected posts above.

BTW I think Borek's method is elegant.

I would imagine a much easier way is to simply

-use the second pH of 6.03 to find Ka2.

-Find the percent ionization, for the second acid it's concentration of H30+ due to the second proton/concentration of H30+ due to the first, this is quite useful in expediting the problem

-use the percent ionization to arrange for an algebraic equation, then simply solve...I'll outline the steps later.

« Last Edit: May 20, 2005, 01:46:49 AM by geodome »

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