Question

# A 0.115 g sample of a diprotic acid of unknown molar mass is dissolved in water...

A 0.115 g sample of a diprotic acid of unknown molar mass is dissolved in water and titrated with 0.1208 M NaOH. The equivalence point is reached after adding 14.8 mL of base.

What is the molar mass of the unknown acid?

Balanced chemical equation is:

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

lets calculate the mol of NaOH

volume , V = 14.8 mL

= 1.48*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 0.1208*1.48*10^-2

= 1.788*10^-3 mol

According to balanced equation

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

= (1/2)*1.788*10^-3

= 8.939*10^-4 mol

This is number of moles of H2A

mass(H2A)= 0.115 g

use:

number of mol = mass / molar mass

8.939*10^-4 mol = (0.115 g)/molar mass

molar mass = 1.286*10^2 g/mol

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