Question

A 0.554 g sample of a diprotic acid is dissolved in water and titrated with 0.150 M NaOH. What is the molar mass of the acid if 36.7 mL of the NaOH solution is required to neutralize the sample? Assume the volume of NaOH corresponds to the second equivalence point.

Answer 1

The general reaction of acid and base is as follows:

H2A + 2 NaOH = 2 NaA + 2 H2O

Given that

0.150 M NaOH

Volume = 36.7 mL of the NaOH

Number of moles = molarity * volume in L

= 0.150 * 36.7/1000

= 0.005505 Moles NaOH

Now calculate the moles of acid as follows:

0.005505 Moles NaOH* 1 mole H1A/2 mole NaOH

= 2.7525*10^-3 moles H2A

Molar mass = amount in g / number of moles

= 0.554 g/ 2.7525*10^-3 moles H2A

=201.27 g/ mole