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.
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
we know number of moles of substance = mass/molar mass
molar mass = mass /number of moles = 0.6071/0.004955 = 122.522 g/mol
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