Question

A 0.401 g sample of a diprotic acid is dissolved in water and titrated with 0.130 M NaOH. What is the molar mass of the acid if 31.3 mL of the NaOH solution is required to neutralize the sample? Assume the volume of NaOH corresponds to the second equivalence point. molar mass: g/mol

Answer 1

Balanced chemical equation is:

2 NaOH + H2A ---> Na2A + 2 H2O

lets calculate the mol of NaOH

volume , V = 31.3 mL

= 3.13*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 0.13*3.13*10^-2

= 4.069*10^-3 mol

According to balanced equation

mol of H2A reacted = (1/2)* moles of NaOH

= (1/2)*4.069*10^-3

= 2.035*10^-3 mol

This is number of moles of H2A

mass(H2A)= 0.401 g

use:

number of mol = mass / molar mass

2.035*10^-3 mol = (0.401 g)/molar mass

molar mass = 1.971*10^2 g/mol

Answer: 197 g/mol