Question

vinegar contains acetic acid. you titrate a vinegar sample with naoh of known concentration according to the following reaction:
$ce{hc_{2}h_{3}o_{2} + naoh -> nac_{2}h_{3}o_{2} + h_{2}o}$
you find that a 25.00 ml sample of vinegar requires 28.33 ml of 0.953 m naoh for titration to the equivalence point. what is the concentration of the acetic acid in the vinegar sample?
$?$ m $ce{hc_{2}h_{3}o_{2}}$

Explanation:

Step1: Recall the titration formula

In acid - base titration, at the equivalence point, the moles of acid are equal to the moles of base (for a 1:1 reaction like this one). The formula for moles of a substance in a solution is \(n = M\times V\), where \(n\) is the number of moles, \(M\) is the molarity, and \(V\) is the volume in liters. For the reaction \(HC_{2}H_{3}O_{2}+NaOH
ightarrow NaC_{2}H_{3}O_{2}+H_{2}O\), the mole ratio of \(HC_{2}H_{3}O_{2}\) to \(NaOH\) is \(1:1\), so \(n_{HC_{2}H_{3}O_{2}}=n_{NaOH}\).

Step2: Calculate moles of NaOH

First, convert the volume of NaOH from milliliters to liters. \(V_{NaOH}=28.33\space mL = 28.33\times10^{- 3}\space L\). The molarity of NaOH, \(M_{NaOH} = 0.953\space M\). Using the formula \(n = M\times V\), the moles of NaOH, \(n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.953\space mol/L\times28.33\times 10^{-3}\space L\).
\[n_{NaOH}=0.953\times28.33\times10^{-3}=0.953\times0.02833\approx0.02699\space mol\]

Step3: Calculate moles of \(HC_{2}H_{3}O_{2}\)

Since the mole ratio of \(HC_{2}H_{3}O_{2}\) to \(NaOH\) is \(1:1\), \(n_{HC_{2}H_{3}O_{2}}=n_{NaOH}\approx0.02699\space mol\).

Step4: Calculate molarity of \(HC_{2}H_{3}O_{2}\)

The volume of the vinegar sample (volume of \(HC_{2}H_{3}O_{2}\) solution) \(V_{HC_{2}H_{3}O_{2}} = 25.00\space mL=25.00\times10^{-3}\space L = 0.025\space L\). The molarity of \(HC_{2}H_{3}O_{2}\), \(M_{HC_{2}H_{3}O_{2}}=\frac{n_{HC_{2}H_{3}O_{2}}}{V_{HC_{2}H_{3}O_{2}}}\). Substituting the values of \(n_{HC_{2}H_{3}O_{2}}\) and \(V_{HC_{2}H_{3}O_{2}}\), we get:
\[M_{HC_{2}H_{3}O_{2}}=\frac{0.02699\space mol}{0.025\space L}\approx1.080\space M\]

Answer:

\(1.08\) (or more precisely \(1.080\))