Answer to Question #56207 in General Chemistry for jocelyn

Question

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.

Answer 1

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:

w(%) = m(HAc)/1000 × 100% = 4.67%