Question

An unknown strong acid 0.344 g with 2 acidic hydrogens (diprotic) is titrated with 35.12 mL of a 0.200 mol/L KOH solution. What is the molar mass of this acid?

Answer 1

Molarity of KOH = 0.200 mol/L

Volume of KOH needed for neutralisation = 35.12 mL = 0.03512 L

No of moles of KOH = 0.200 mol/L × 0.03512 L = 0.007024 mol

Let the diprotic acid be H₂X

The neutralisation reaction is given by -

H₂X + 2 KOH → K₂X+ 2 H₂O

2 mol NaOH reacts with 1 mol H₂X

0.007024 mol NaOH reacts with (1 mol × 0.007024 mol) / 2 mol = 0.003512 mol of H₂X.

Now,

No of moles = mass given / molar mass

Molar mass = mass given / no of moles

Molar mass = 0.344 g / 0.003512 mol = 97.95 g/mol

Answer : 97.95 g/mol

Note : If you take the round figure it becomes 98 g/mol which is the molar mass of sulphuric acid.

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