Asample of 0.2140 g of an unknown monoprotic acid was dissolved in 25.0 ml of water and titrated with 0.0950mnaoh. the acid required 27.4 ml of base to reach the equivalence point. what is the molar mass of the acid?

in a monoprotic acid , its capable of dissociating 1 H⁺ ion . The base used is NaOH which is capable of giving out 1 OH⁻ ion.

Therefore both base and acid react in a 1:1 molar ratio, that means for each 1 mol of monoprotic acid there is , it requires 1 mol of NaOH to be neutralised.

So at equivalence point the number of acid moles present should react with an equal number of NaOH moles to be neutralised

number of NaOH moles added - concentration x volume

number of NaOH moles - 0.0950 mol/dm³ x 27.4 x 10⁻³ dm³ = 0.00260 mol

since molar ratio of NaOH to Acid is 1:1

number of acid moles present is also - 0.00260 mol

mass of acid added - 0.2140 g

we can find molar mass using following equation

molar mass = mass / number of moles

molar mass = 0.2140 g / 0.00260 mol = 82.3 g/mol

therefore molar mass of monoprotic acid is 82.3 g/mol

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