Question

4. An monoprotic acid was isolated from an organic chemistry reaction. Titration of a sample of pure acid (1.2131 g) with NaOH (0.0944 M) to a phenolphthalein endpoint required 41.21 mL of base. What is the molecular weight of the unknown acid?

Answer 1

The general reaction of acid and base is as follows:

HA + NaOH = Na A + H2O

Number of mole s= molarity * volume in L

= 0.0944 mole / L * 41.21ml *1 L /1000 ML

= 0.00389 Moles NaOH

Now calculate the moles of acid as follows:

0.00389 Moles NaOH *1 mole HA /1 Moles NaOH

=0.00389 Moles HA

Molecular mass = amount in g / number of moles

= 1.2131 g / 0.00389 Moles

=311.85 g / mole