An analytical chemist weighs out 0.463g of an unknown triprotic acid into a 250mL volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with 0.1800M NaOH solution. When the titration reaches the equivalence point, the chemist finds he has added 40.2mL of NaOH solution. Calculate the molar mass of the unknown acid.

Question

aaronq · Answer

how many moles of base did you use?

anonymous · Answer

40.2 mL * (1L/1000 mL) * 0.18 M NaOH = 0.007236 moles of NaOH (base) were used.

aaronq · Answer

so, you know those were the moles of acid present. Though there was 3 times as much because each acid molecule provided 3 protons. 

$H_3A \rightarrow A^{3-}+ 3H^+$

so divide the moles of NaOH by 3 and that's how many moles of acid you had.

then use the mass (given) to find the molar mass