posted by Holly .
A solution contains 25 mmol of H3PO4 and 10. mmol of NaH2PO4. What volume of 2.0 M NaOH must be added to reach the second equivalence point of the titration of the H3PO4 with NaOH?
Look at titrating the first H, then double it.
H3PO4 + NaOH ==> NaH2PO4 + H2O
mols H3PO4 = 25 mmols.
So you will need 25 mmols NaOH to react with the first H of H3PO4.
M NaOH = mols NaOH/L NaOH
2.0M = 25 mmols/mL
mL = 25/2.0 = 12.5 mL.
So it will take 25 to neutralize the second H of H3PO4 (and 37.5 mL to neutralize the 3rd H although you probably wouldn't know when you were there.)