A sample of a monoprotic acid (HA) weighing 0.384 g is dissolved in water and the solution is titrated with aqueous NaOH. If 30.0 mL of 0.100 M NaOH is required to reach the equivalence point, what is the molar mass of HA?

(a) 211 g/mol
(b) 128 g/mol
(c) 81.0 g/mol
(d) 37.0 g/mol
(e) 20.3 g/mol

The correct answer is option b.

Explanation:

Moles of NaOH = n

Volume of NaOH solution = 30.0 mL =0.030 L ( 1mL =0.001 L)

Molarity of the NaOH solution = 0.100 M

According to reaction , 1 mol of NaOH reacts with 1 mole of HA, then, 0.00300 moles of NaOH will react with :

of HA

Moles of HA = 0.00300 mol

Mass of HA = 0.384 g

Molar mass of HA = M

Mass of 0.00300 moles of HA = 0.384 g

128 g/mol is the molar mass of HA.