Question

A 0.350-g sample pf an acid, HX, is dissolved in 25.00 mL of water. The resulting solution is titrated with 0.140 M NaOH solution, and 25.40 mL of the base solution is required to neutralize the acid. Calculate the molar mass of the acid.

Answer 1

HX + NaOH ==> H₂O + NaX

1 mol HX reacts with 1 mol NaOH

moles NaOH = 25.40 mL x 10^-3 L x 0.140 M = 0.003556 mol

same number of moles of acid will be exactly required to neutralize = 0.003556 mol

molar mass = mass of acid / moles of acid = 0.350 g / 0.003556 mol = 98.4251968504 g/mol

Answer: the molar mass of the acid = 98.42 g/mol (4 significant figures answer)

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