Question

1. The student dissolved 2.050 grams of acid in 50 mL of distilled water. He then titrated the unknown acid with a standardized NaOH solution. Unfortunately, the student over shot the end point and had to back titrate the solution with 2.17 mL of 0.98861 M HCl. Once the final end point was determined, the student had added a total of 29.15 mL of a 1.0324 M NaOH solution.

A) How many moles of OH^- were used to reach the end point?

B) How many moles of H⁺ from HCl were used for back titration?

C) How many moles of H⁺ from the solid were neutralized?

D) What is the equivalent mass of the unknown acid?

Answer 1

A) No of mole of OH- used to reach the end point = moles of NaOH added - moles of HCl used to back titration.

   = 29.15*1.0324 - (2.17*0.98861)

   = 27.95 mmole

   = 0.02795 mole

B) No of mole of H+ were used to back titration = 2.17*0.98861

   = 2.1453 mmole

   = 0.0021453 mole

C) No of moles of H+ neutralised from solid = No of moleS of OH- added to reach endpoint

No of moles of H+ neutralised from solid = 0.02795 mole

D) Equivalent mass of the unknown acid = w/n

       = 2.050 / 0.02795

       = 73.35