in the titration of 79.0 ml of 0.400 m hcooh with 0.150 m lioh, how many ml of lioh are required to reach the equivalence point?

Answer

210.67 ml

Calculate the number of moles of HCOOH: n(HCOOH) = V(HCOOH) x C(HCOOH) = 79.0 ml x 0.400 mol/L = 31.60 mmol.
Use the balanced chemical equation for the reaction between HCOOH and LiOH to determine the number of moles of LiOH required at the equivalence point: HCOOH + LiOH -> LiCOOH + H2O. The stoichiometry of the reaction is 1:1, so n(LiOH) = n(HCOOH) = 31.60 mmol.
Calculate the volume of LiOH required at the equivalence point: V(LiOH) = n(LiOH) / C(LiOH) = 31.60 mmol / 0.150 mol/L = 210.67 ml.