in a typical titration a student needed 36.48mL OF 0.1067 M NaOH to reach the phenolphthalein endpoint. if the volume of the original vinegar sample was 5.00mL calculate the mass % of acetic acid in the vinegar. Report your answer to the appropriate number of significant figures.
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Expert's answer
2015-11-09T08:28:59-0500
The amount of acetic acid in 5.00 ml equals: ν(HAc) = ν(NaOH) = C(NaOH)×V(NaOH) = 0.1067 mol/L × 0.03648 ml = 3.8924 mmol The mass of acetic acid is: m(HAc) = ν(HAc)×M, where M - the molar weight of acetic acid. m(HAc) = 3.8924 mmol × 60 g/mol = 233.54 mg (in 5 ml). In 1 L the mass of acetic acid is: m(HAc) = (1000/5)× 233.54 mg = 46.71 g Thus, the mass % of acetic acid is determined:
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