The first eq point is at 13.32 mL and that titrates the first H of the H3PO4. If the sample contained ONLY H3PO4, then the second eq point would be at 13.32 x 2 or 26.64 mL to titrate the second H. So the difference between 28.65 and 26.64 must be due to the H2PO4^-.
(28.65-26.64/1000) x 0.1476 M = moles H2PO4^- and that is the moles in 100 mL sample. That times 10 will be the moles in L.
The phosphoric acid in a 100.00 mL sample of a cola drink was titrated with 0.1476 M NaOH. The first equivalence point was detected after 13.32 mL of base added, and the second equivalence point after 28.65 mL. Calculate the concentration of H2PO4- in mol/L. (Hint: if only H3PO4 were present, where would the second equivalence point be?)
I have tried using C1V1 = C2V2, using as many different combination of volumes as I can, but I can't get the answer ( I know what the answer is supposed to be and cannot come up with it)... I think I am missing something that's hidden within the hint but I don't know what that could be...any help would be appreciated.
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