determining the volume of base needed to titrate

Question

A chemistry student weighs out 0.385g of citric acid (H3C6H5O7), a triprotic acid, into a 250.mL

 volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1500M NaOH

 solution. Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Round your answer to 3 significant digits.

Answer 1

Since citric acid is a triprotic acid, we can represent it as H₃A (3 titrate-able) protons. When titrated with NaOH, the reaction would look like this ...

H₃A + 3NaOH ==> Na₃A + 3H₂O (note the ratio of 3 mols NaOH for each mole citric acid)

molar mass citric acid (H₃C₆H₅O₇) = 191.12 g/mol

moles of citric acid present = 0.385 g x 1 mol / 191.12 g = 0.00201 mols

moles NaOH needed = 0.0201 mols citric acid x 3 mols NaOH / mol citric acid = 0.00604 mols NaOH needed

volume of 0.150 M NaOH = 0.00604 mols NaOH x 1 L / 0.1500 mols = 0.0403 L = 40.3 mls (3 sig. figs.)