Question

1. A student measures out 0.120 g of maleic acid (C₄H₄O₄) and titrates it with 0.100 M NaOH. How many mL of NaOH will be added to reach the second equivalence point? Show your calculations. What would be the volume of NaOH added at the first equivalence point?

Answer 1

CH - coon + 2NaOH Cu-coo-Nat CH-Coo-Nat CH - cool maleic acid (diprotic acid) molar mass = 11619 mol Given weight of maleic molar man = 116.19 acid = 0.120g hana mol mol q maleic acid = 0.1209_ = 0.00103 mol 116.1 & moll

At second equival auld I Second equivalence point means whole mallu completely neutralised by W2044 2 mol y Naon completely reset with I mol maleic aud or Imol maleic acid required amol Naon 0.00103 mol maleic acid required = 0.00206 - moly Naon needed = 0.00206 Molarity = 0.100M volume of Naoh= mol of Naon molaring ___ = 0.00206 mot = 0.02064 0.100 olut II L = 1000mLt ! 0.02062- 206ml so 20 6 mL g Naon needed to reach second equivalence point: At first epuivalena point half maleic acid neutralised. so volume needed to reach first equivalence = 10.3mL point