Astudent completes a titration of an unknown monoproticacid. in this experiment, 0.79 g of the acid is dissolved in 250.0 ml of water. it requires 13.48ml of 1.0 m naoh to reach theequivalence point. what is the molar mass of the acid?

58.61 g/mol

Explanation:

The acid us monopromatic and hence, 1 mole of it is required to react with  mole of NaOH in order to reach the equivalence point.

In order to detrmine the concentration of the acid:

, where

Ca = concentration of acid, Cb = concentration of base (1.0), Va = volume of acid (250 mL), Vb = volume of base (13.48 mL), na = number of moles of acid and (1) nb = number of moles of base (1).

=

      = 0.05392 M

Concentration = mole/volume

0.05392 = mole/0.25

mole of acid = 0.01348

Also, mole = mass/molar mass

Molar mass = mass/mole

                      = 0.79/0.01348

                       = 58.61 g/mol

The molar mass of the acid is 58.61 g/mol.

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