First off the ratio is 1:1 ratio according to the equation.
According to the titration graph for trial 1 the first equivalence point is 24.5 ml at 4.1 ph and the 2nd equivalence point was 38.0ml at 9.1 ph. Also, the molarity of NaOH was standardized at .095375M.
.095375M X 24.5 = molarity acid X 100ml=.023367M X 2 =.046734
.095375M X 38.0ml= molarity acid X 100ml=.036243M X 2=.072485
What is neutralized at first equivalence point?
the total neutralization of HCl to NaCL and H3PO4 to NaH2PO4 according to
HCl(aq) + NaOH(aq)>H2O(l)+NaCl(aq)
the second equivalence point is the neutralization of NaH2PO4(aq)+H2O(l)
The equation for the standardizeof NaOH with potassium hydrogen phthalate