Question 10.17: When a 5.00 mL sample of household vinegar (dilute aqueous a...
When a 5.00 mL sample of household vinegar (dilute aqueous acetic acid) is titrated, 44.5 mL of 0.100 M NaOH solution is required to reach the end point. What is the acid concentration of the vinegar in moles per liter, equivalents per liter, and milliequivalents per liter? The neutralization reaction is
CH_3CO_2H(aq) + NaOH(aq) \longrightarrow CH_3CO_2 ^- Na^+(aq) + H_2O(l)ANALYSIS To find the molarity of the vinegar, we need to know the number of moles of acetic acid dissolved in the 5.00 mL sample. Following a flow diagram similar to Figure 10.9, we use the volume and molarity of NaOH to find the number of moles. From the chemical equation, we use the mole ratio to find the number of moles of acid, and then divide by the volume of the acid solution. Because acetic acid is a monoprotic acid, the normality of the solution is numerically the same as its molarity.
BALLPARK ESTIMATE The 5.00 mL of vinegar required nearly nine times as much NaOH solution (44.5 mL) for complete reaction. Since the neutralization stoichiometry is 1:1, the molarity of the acetic acid in the vinegar must be nine times greater than the molarity of NaOH, or 0.90 M.
Substitute the known information and appropriate conversion factors into the flow diagram, and solve for the molarity of the acetic acid:
(44.5 \cancel{mL NaOH}) \left(\frac{0.100 \cancel{mol NaOH}}{1000 \cancel{mL}} \right) \left(\frac{1 mol CH_3CO_2H}{ 1 \cancel{mol NaOH}} \right) \\ \hspace{30 pt} \hspace{30 pt} \times \left(\frac{1}{0.005 00 L} \right) = 0.890 M CH_3CO_2H \\ \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} = 0.890 N CH_3CO_2HExpressed in milliequivalents, this concentration is
\frac{0.890 Eq}{L } \times \frac{1000 mEq}{1 Eq} = 890 mEq/LBALLPARK CHECK The calculated result (0.890 M) is very close to our estimate of 0.90 M.