1) A student dissolved 0.1916 g of an unknown diprotic acid in 100 mL of distill

Question

1) A student dissolved 0.1916 g of an unknown diprotic acid in 100 mL of distilled water. The acid was then titrated with 0.1025M NaOH solution. The second equivalence point showed the sharpest change in pH, and so it was used to determine the molar mass of the unknown acid. The volume of NaOH needed to reach the equivalence point was 27.5 mL.

a. Calculate the number of moles of NaOH used in the titration to reach the second equivalence point.

b. Calculate the number of moles of diprotic acid, based on the fact that we are examining the second equivalence point.

c. Calculate the molar mass of the diprotic acid.

Explanation / Answer

a) moles of NaOH = MV = 0.1025 x 27.5/1000 = 0.00282

b) moles of diprotic acid = 0.00282/2 = 0.00141 ( acid is diprotic and hence two equiavelnts points and hence moles of acid = moles of base used for second equivalence point /2 )

c) molar mass = mass/moles = 0.1916/0.00141 = 135.9 gm/mol

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.