Questions: An analytical chemist weighs out 0.115 g of an unknown monoprotic acid into a 250 mL volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with 0.1500 M NaOH solution. When the titration reaches the equivalence point, the chemist finds he has added 7.9 mL of NaOH solution. Calculate the molar mass of the unknown acid. Be sure your answer has the correct number of significant digits.

Solution Steps

The moles of NaOH used in the titration can be calculated using the given volume and molarity:

Moles of NaOH = Volume (L) * Molarity (mol/L)

Moles of NaOH = (7.9 mL * (1 L / 1000 mL)) * 0.1500 mol/L

Moles of NaOH = 0.001185 mol

Since the acid is monoprotic, the mole ratio between the acid and NaOH is 1:1 at the equivalence point. Therefore, the moles of the unknown acid are equal to the moles of NaOH.

Moles of unknown acid = 0.001185 mol

The molar mass of the unknown acid can be calculated using the given mass and the calculated moles:

Molar mass = Mass (g) / Moles

Molar mass = 0.115 g / 0.001185 mol

Molar mass = 97.0 g/mol

Final Answer: The molar mass of the unknown acid is 97.0 g/mol.