Question

A 0.3012 g sample of an unknown monoprotic acid requires 24.13 mL of 0.0944 M NaOH for neutralization to a phenolphthalein end point. There are 0.32 mL of 0.0997 M HCl used for back-titration.

How many moles of OH- are used? How many moles of H+ from HCl?

How many moles of H+ are there in the solid acid? (Use Eq. 5.) moles H+ in solid iacid =moles OH- in NaOH soln. - moles H+ in HCl soln.

What is the molar mass of the unknown acid? (Use Equation 4) MM = grams of acid/moles of H+ ion furnished

Answer 1

NaOH is made to react with an unknown acid, and the remaining NaOH is made to react with HCl. Therefore

Moles of unknown acid = mols of NaOH - Moles of HCl

Moles of NaOH = Volume X molarity = 24.13mL X 0.0944 = 2.28mmoles

Moles of HCl = 0.32 X 0.0997 = 0.032 mmoles

Hence mole of Unknown acid = 2.28mmoles - 0.032 mmoles = 2.25mmoles

(A) Moles of OH- consumed by unknown acid = 2.25mmoles. Moles of H+ (from HCl) consumed = 0.032 mmoles

(B) Moles of H+ in the solid acid (UNKNOWN ACID) = 2.25 mmoles

(C) molar mass = mass in grams / no. of moles

molar mass = 0.3012 / 2.25 X 10^-3 = 133.86 g/mol

HINT: 1 mmol = 10^-3 moles