A 5.00g quantity of a diprotic acid was dissolved in water and made up exactly 250 mL. Calculate the molar mass if the acid is 25.0 mL of this solution required 11.1 mL of 1.00 KOH for neutralization. Assume both protons of the acid were titrated.

The molar mass of the diprotic acid is 90.10 g/mol.

Explanation:

The acid is 25.0 mL of this solution required 11.1 mL of 1.00 KOH for neutralization.

To calculate the concentration of acid, we use the equation given by neutralization reaction:

 ( neutralization )

where,

are the n-factor, molarity and volume of diprotic acid

are the n-factor, molarity and volume of base which is KOH.

We are given:

Putting values in above equation, we get:

Molarity of acid solution = 0.222 M =0.222 mol/L

Volume of original solution = 250 mL = 0.250 L ( 1 mL = 0.001 L)

Moles of diprotic acid in 0.250 L solution :

Mass of diprotic acid = m = 5.00 g

The molar mass of the diprotic acid is 90.10 g/mol.

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