Question

A solution of an unknown monoprotic acid was prepared by dissolving 72.9 mg of the acid into 50. mL of water. 10.0 mL of 0.10 M NaOH is needed to titrate a 25 mL aliquot of the unknown monoprotic acid. From this info, calculate the molar mass of the unknown acid (g/mol).

Answer 1

36.45g/mol

Explanation

Reaction between NaOH and monoprotic acid(HA) is

NaOH + HA ----> NaA + H2O

1:1 molar reaction

molarity = number of moles of solute per liter of solution

moles of NaOH consumed = (0.10mol/1000ml)×10ml = 0.001mol

0.001mol of NaOH react with 0.001moles of HA

number of moles in 25ml aliquot = 0.001mol

number of moles of HA in 50ml solution = 2× 0.001mol = 0.002

molar mass = mass /number of moles

molar mass of HA = 0.0729g/0.002mol = 36.45 g/mol