Question

Potassium hydrogen phthalate is a monoprotic acid. Dihydrogen phthalate (H2C8H4O4) is a diprotic acid. An analyte sample contains 0.127 M KHP and 0.0678 M H2C8H4O4. What volume of a 0.205 M NaOH solution is required to neutralize 25.0 mL of the analyte solution?

Molar masses:
H2C8H4O4 - 166.13 g/mol
KHP - 204.23 g/mol

Answer 1

I think something is missing but I don't know what.
KHP + NaOH ==> H2O + NaKP
H2P + 2NaOH ==> 2H2O + Na2P
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Suppose the sample is 100% KHP, then
25.0 mL x 0.127M = 3.175 mmols.
That will take 3.175 mmols NaOH
M NaOH = mmols/mL; rearrange to
mL = mmols NaOH/M NaOH= 3.175/0.205 = about 15 mL of the NaOH. That's one possible solution.

Suppose the sample is 100% H2P, then
25.0 mL x 0.0678M = 1.695 mmols H2P.
mmols NaOH = 2 x 1.695 = 3.39 and
3.39/0.205 = about 16 mL of the NaOH. That's a second possible solution.

Doesn't it stand to reason that for any possible combination there will be a solution; therefore, there is no unique solution.