To find the molarity of the vinegar solution, let's go through the steps carefully:
1. Initial Volume of Vinegar Solution ([tex]$V_i$[/tex]):
We start with 5.00 mL of vinegar solution.
2. Final Volume of Solution ([tex]$V_f$[/tex]):
The final volume of the solution after titration is recorded as 16.67 mL.
3. Volume of NaOH Titrant Used ([tex]$V_{\text{titrant}}$[/tex]):
The volume of NaOH titrant used during the titration is the difference between the final volume and the initial volume:
[tex]\[
V_{\text{titrant}} = V_f - V_i = 16.67 \, \text{mL} - 5.00 \, \text{mL} = 11.67 \, \text{mL}
\][/tex]
4. Molarity of NaOH ([tex]$M_{\text{NaOH}}$[/tex]):
The molarity of the NaOH solution used for titration is given as 0.245 M.
5. Moles of NaOH Used:
To find the moles of NaOH used, we use the volume of NaOH solution (in liters) and its molarity:
[tex]\[
\text{Moles of NaOH} = V_{\text{titrant}} \times M_{\text{NaOH}} = 11.67 \, \text{mL} \times 0.245 \, \text{M} \, \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.00285915 \, \text{moles}
\][/tex]
6. Stoichiometry of the Reaction:
The balanced chemical equation shows a 1:1 mole ratio between acetic acid (HC₂H₃O₂) and sodium hydroxide (NaOH). Thus, the moles of acetic acid (HC₂H₃O₂) used are the same as the moles of NaOH:
[tex]\[
\text{Moles of HC}_2\text{H}_3\text{O}_2 = 0.00285915 \, \text{moles}
\][/tex]
7. Molarity of Vinegar Solution:
Finally, we determine the molarity of the vinegar solution by dividing the moles of acetic acid by the initial volume of the vinegar solution (in liters):
[tex]\[
\text{Molarity of Vinegar} = \frac{\text{Moles of HC}_2\text{H}_3\text{O}_2}{V_i \, \times \frac{1}{1000}} = \frac{0.00285915 \, \text{moles}}{5.00 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}}} = 0.57183 \, \text{M}
\][/tex]
Therefore, the molarity of the vinegar solution for this trial is [tex]\(0.57183\)[/tex]. Enter this as 0.5718300000000001 without units.
