Question

A student completes a titration of an unknown diprotic acid. In this experiment, 0.79 g of the acid is dissolved in 250.0 mL of water. It requires 13.48 mL of 1.0 M NaOH to reach the second equivalence point. What is the molar mass of the acid?

Answer 1

117.21g/mol

Step-by-step explanation:

The reaction of a diprotic acid, H₂X with NaOH occurs as follows:

H₂X + 2NaOH → Na₂X + 2H₂O

Where 2 moles of NaOH reacts per mole of the diprotic acid

Knowing the moles of NaOH and the chemical equation we can determine the moles of the diprotic acid. With the moles and its mass we can find the molar mass of the acid, thus:

Moles of the diprotic acid:

To reach the equivalence point, it requires 13.48mL = 0.01348L of 1.0M NaOH. That is:

0.01348L × (1mol NaOH / L) = 0.01348 moles NaOH

As 2 moles of NaOH reacts per mole of the diprotic acid, moles of the acid are:

0.01348moles NaOH × (1mol H₂X / 2mol NaOH) = 0.00674 moles of H₂X

Molar mass:

The molar mass of a compound is the ratio between the mass you have per mole. As mass is 0.79g and moles are 0.00674moles, molar mass is:

Molar mass: 0.79g / 0.00674mol =

117.21g/mol