Question
A 12.5 mL sample of vinegar, containing acetic acid, was titrated using $0.504 \mathrm{M} \mathrm{NaOH}$ solution. The titration required $20.65 \mathrm{~mL}$ of the base. What was the molar concentration of acetic acid in the vinegar?
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5 mL sample of vinegar containing acetic acid was titrated using a 0.504 M NaOH solution. The titration required 20.65 mL of the base. We are asked to find the molar concentration of acetic acid in the vinegar. Show more…
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A 12.5 mL sample of vinegar, containing acetic acid, was titrated using 0.504 M NaOH solution. The titration required 20.65 mL of the base. What was the molar concentration of acetic acid in the vinegar?
A $12.5 \mathrm{~mL}$ sample of vinegar, containing acetic acid, was titrated using $0.504 M \mathrm{NaOH}$ solution. The titration required $20.65 \mathrm{~mL}$ of the base. (a) What was the molar concentration of acetic acid in the vinegar? (b) Assuming the density of the vinegar is $1.01 \mathrm{~g} \mathrm{~mL}^{-1}$, what was the percent (by mass) of acetic acid in the vinegar?
A $25.0 \mathrm{~mL}$ sample of vinegar with a density of $1.01 \mathrm{~g}$ $\mathrm{mL}^{-1}$, containing acetic acid, was titrated using $0.504 \mathrm{M}$ $\mathrm{NaOH}$ solution. The titration required $42.54 \mathrm{~mL}$ of the base. What was the molar concentration and the percent by mass of acetic acid in the vinegar?
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