Question

A sample of 0.2140 grams of an unknown monoprotic weak acid was dissolved in 25.0mL of water and titrated with 0.0950M NaOH. The acid required 15.50mL of NaOH to reach the equivalence point. What is the molar mass of the unknown acid?

Answer 1

one mole mono protic acid is neutralized by one mole NaOH.

mole of NaOH = volume of NaOH solution in L * molarity = 0.01550 L * 0.0950 mole / L = 1.47 * 10^-3 mole.

thus

mole of mono protic acid = 1.47 * 10^-3 mole

we know,

mole = mass / molar mass

or

molar mass = 0.2140 g / 1.47 * 10^-3 mole = 145.33 g / mole.

molar mass of the unknown acid = 145.33 g / mole.