Question

A 0.6071 g sample of a weak monoprotic acid was titrated to its endpoint using 0.1325 M NaOH. The initial volume of NaOH was 0.58 mL. The final volume of NaOH was 37.98 mL. What is the molar mass of the weak monoprotic acid? Show all work.

Answer 1

Volume of NaOH used for this titration = 37.98-0.58 = 37.4 mL

the concentration of NaOH = 0.1325 M

since Molarity is the number of moles of solute in 1000 mL solution, we can calculate the number of moles present in 37.4 mL = (0.1325/1000) x 37.4 = 0.004955 M

Since Acid as well as NaOH are monoprotic, the number of moles of acid present is = 0.004955 moles

also,

we know number of moles of substance = mass/molar mass

or

molar mass = mass /number of moles = 0.6071/0.004955 = 122.522 g/mol