Question

An analytical chemist weighs out 0.093g of an unknown monoprotic acid into a 250mL volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with 0.1600M NaOH solution. When the titration reaches the equivalence point, the chemist finds he has added 6.5mL of NaOH solution. Calculate the molar mass of the unknown acid. Round your answer to 2 significant digits.

Answer 1

Answer – Given, mass of monoprotic acid = 0.093 g , volume of monoprotic acid = 250 mL , [NaOH] = 0.1600 M , volume of NaOH required at reaches the equivalence point = 6.5 mL.

First we need to calculate the moles of NaOH

Moles of NaOH = molarity * volume (L)

                            = 0.1600 M * 0.0065 L

                            = 0.00104 moles

Assume monoprotic acid is HA

We know monoprotic acid has 1 H⁺ and NaOH is also monoprotic base, means it has 1 OH^-, so reaction is like

HA + NaOH -----> NaA + H₂O

From the above balanced equation

1 moles of NaOH = 1 moles of HA

So, 0.00104 moles of NaOH = 0.00104 moles of HA

So, molar mass of unknown acid = mass / moles

                                                   = 0.093 g / 0.00104 moles

                                                     = 89.42 g/mol