Question

5.0 mL of commercial vinegar were taken and diluted to 500 mL with distilled water. From this freshly prepared solution, 25.0 mL were taken and titrated with a 0.1093 M solution of sodium hydroxide. If 5.71 mL of the sodium hydroxide solution were consumed in this titration, find the percent by mass of acetic acid in commercial vinegar. (Assume the density of commercial vinegar is equal to that of water)

Answer 1

Moles of acetic acid in vinegar sample = Moles of sodium hydroxide used in titration

C(NaOH) = 0.1093 M = 0.1093 mol/L

Proportion 1:

0.1093 mol – 1000 mL

x mol – 5.71 mL

x $= \frac{0.1093 \times 5.71}{1000} = 0.6241 \times 10^{-3}\; mol$

Moles of acetic acid in vinegar sample = Moles of sodium hydroxide used in titration

n(acetic acid) $= 0.6241 \times 10^{-3} \;mol$ (in 25 mL of sample for titration)

Proportion 2:

$0.6241 \times 10^{-3} mol$ – 25 mL

y mol – 500 mL

y $= \frac{0.6241 \times 10^{-3} \times 500}{25} = 12.48 \times 10^{-3} \;mol$ (in 500 mL)

n(acetic acid) in 500 mL = n(acetic acid) in 5.0 mL $= 12.48 \times 10^{-3} \;mol$

M(acetic acid) = 60 g/mol

m(acetic acid) $= n \times M = 12.48 \times 10^{-3} \times 60 = 0.7488 \;g$

Proportion 3:

0.7488 g – 5 mL

z g – 100 mL

z = 14.97 g or 15 % (in 100 mL)

Answer: 15 %