A chemistry student weighs out 0.0349g of formic acid HCHO2 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 equivalence point. Round your answer to 3 significant digits.

the molar mass of formic acid CH2O2 = 46.03 g/mol

the number of moles of HCHO2 = 0.0349 g / 46.03 g/mol

n = 7.58 × 10^-4 mol

according to the balanced chemical equation

HCHO2 + NaOH → H2O + NaCHO2

1 mol of formic acid will react with 1 mol of NaOH

so we need 7.58 × 10^-4 mol of NaOH

the molarity = 0.1500M = 7.58 × 10^-4 mol / volume

the volume = 7.58 × 10^-4 mol / 0.1500M

= 5.05 × 10^-3 L = 5.05 ml

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