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A chemistry student weighs out 0.0432 g of formic acid (HCHO2) into a 250 mL volumetric...

A chemistry student weighs out 0.0432 g 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 0.0700 M NaOH solution. Calculate the volume of NaOH solution that the student will need to add to reach the equivlence point. Be sure your answer has the correct number of significant digits.

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Answer #1

HCHO2(aq) + NaOH(aq) ---------------> NaCHO2(aq) + H2O(l)

no of moles of formic acid   = W/G.M.Wt

                                             = 0.0432/46   = 0.00094moles

1 mole of HCHO2 react with 1 mole of NaOH

0.00094 moles of HCHO2 react with 0.00094 moles of NaOH

no of moles of NaOH   = molarity * volume in L

0.00094                       = 0.07 * volume in L

volume in L                 = 0.00094/0.07   = 0.0134L   = 13.4ml

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