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?
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CH3COOH + NaOH -> CH3COONa + H2O Show more…
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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 16.0 mL sample of vinegar, which is an aqueous solution of acetic acid, CH3COOH, requires 18.5 mL of 0.400 M NaOH to reach the endpoint in a titration. CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l) What is the molarity of the acetic acid solution?
Sri K.
Acetic acid $\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)$ is an important ingredient of vinegar. $\mathrm{A}$ sample of $50.0 \mathrm{mL}$ of a commercial vinegar is titrated against a $1.00 M \mathrm{NaOH}$ solution. What is the concentration (in $M$ ) of acetic acid present in the vinegar if $5.75 \mathrm{mL}$ of the base is needed for the titration?
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