Question

A 0.521 g sample of a diprotic acid is dissolved in water and titrated with 0.160 M NaOH. What is the molar mass of the acid if 40.4 mL of the NaOH solution is required to neutralize the sample? Assume the volume of NaOH corresponds to the second equivalence point. molar mass: g/mol

Answer 1

H2X + 2NaOH <==> 2H2O + Na2X

1 mol acid requires 2 mol base

moles base = 40.4 mL x 10^-3 L x 0.160 M = 0.006464 mol

moles acid = 0.006464 / 2 = 0.003232 mol

molar mass = mass / moles = 0.521 g / 0.003232 mol = 161.20049505 g/mol

Answer: 161.2 g/mol

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