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.