Let me type the problem and my methods

139)

A 225-mg sample of a diprotic acid is dissolved in enough water to make 250.0mL of solution. The pH of this solution is 2.06. A saturated solution of calcium hydroxide (Ksp- 1.3e-6) is prepared by adding excess calcium hydroxide to pure water and then rmeoving the undissolved solid by filltration. Enough of the calcium hydroxide solution is added to the solution of acid to reach the second equivalence point. The pH at the second equivalence point (as determined by pH meter) is 7.96. The first dissociation constant for acid (Ka1) is 5.90e-2. Assume that the volumes of solutions are additive, all solutions are at 25degC, and that Ka1 is at least 1000 times greater than Ka2.

a) calculate the molar mass of the acid

b. calculate the second dissociation constant for the acid Ka2

my method,

first to calculate molar mass of the acid

to calculate molar mass i set up a table (concentrations)

H_{2}A + H_{2}O <----------> H_{3}O^{+} + HA^{-}

I mm/225 - 0 0

C -x - +x +x

E mm/225-x - x x

b/c pH =2.06 we know [H] = .0087096=x

Ka1= [HA][H]/[H2A] = x^{2}/[(mm-225)-x] mm= 2.24,

please check this... this step is crucial to the rest of my method... right now im iffy if the concentration of H2A really is mm/225... can someone verify or prove differently my proof is

molarity = mmol/mL

mmol= (molar mass)1000

molarity = molarmass(1000)/mL

L=1000mL

molarity = molarmass/L MM/.225 or 1000MM/mL but using this method i get a molar mass < 1 which cant be right... any help?