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).
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
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