Question

1.25 grams of unknown acid was dissolved in 15.00 mL of water and then titrated with 0.100 M NaOH(aq). The

Answer 1

The molar mass of the acid, given the data is 625 g/mole

What is equivalence point?

This is a point is a titration reaction where the amount of the acid added is enough to neutralize the base

This simply implies that the number of mole of the acid used is equal to the number of mole of the base.

How to determine the mole of the base

• Volume of base (NaOH) = 20 mL = 20 / 1000 = 0.02 L
• Molarity of base (NaOH) = 0.1 M
• Mole of base (NaOH) = ?

Mole = molarity × volume

Mole of base (NaOH) = 0.1 × 0.02

Mole of base (NaOH) = 0.002 mole

How to determine the molar mass of the acid

• Mole of base (NaOH) = 0.002 mole
• Mole of acid = Mole of base (NaOH) = 0.002 mole
• Mass of acid = 1.25 g
• Molar mass of acid =?

Molar mass = mass / mole

Molar mass of acid = 1.25 / 0.002

Molar mass of acid = 625 g/mole

Learn more about titration:

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