Question

A .288 g sample of an unknown monoprotic organic acid is dissolved in water and titrated with a .115 M sodium hydroxide solution. After the addition of 17.54 mL of base, a pH of 4.92 is recorded. The equivalence point is reached when a total of 33.83 mL of NaOH is added. What is the molar mass of the organic acid?

Answer 1

A generic equation can be written as follows:
RCOOH + NaOH ==> H2O + RCOONa
mols NaOH = L x M
mols RCOOH = mols NaOH (since the acid is monoprotic).
mols RCOOH = grams/molar mass.
You have mols and grams. Solve for molar mass.
The pH of 4.92 at 17.54 mL NaOH is extraneous information unless you are asked to calculate Ka for the acid.

Answer 2

moles OH- = (liters)(molarity)
(0.115 mol NaOH/L)(0.03383L) = 3.89x10-3 mol OH- used to reach the end point

Moles acid = moles OH-
moles of acid neutralized = 3.89x10-3 mol

0.288 g acid/0.03383L = 8.513g/L
(8.513g/L)/(0.115 mol/L) = ________g/mole

Answer 3

Yeah asking for Ka was the second part of the question, but it wasn't one of the review questions