Final answer:
The molar mass of a monoprotic acid neutralized by NaOH is calculated by finding the moles of NaOH used, assuming a 1:1 reaction ratio, and then using the mass of the acid sample to determine its molar mass, which is found to be 87.77 g/mol.
Step-by-step explanation:
To calculate the molar mass of a monoprotic acid neutralized by sodium hydroxide (NaOH), first, determine the number of moles of NaOH used in the titration. This can be calculated by multiplying the concentration of NaOH by its volume in liters:
# mol NaOH = 14.5 mL × 0.11 M = 0.001595 mol NaOH
Since the acid is monoprotic, it reacts with NaOH in a 1:1 molar ratio, the number of moles of acid is equal to the number of moles of NaOH. Now, calculate the molar mass of the acid using the mass of the acid sample and the number of moles:
Molar mass of acid = Mass of acid / Number of moles of acid
Molar mass of acid = 0.14 g / 0.001595 mol = 87.77 g/mol
Therefore, the molar mass of the monoprotic acid is 87.77 g/mol.