To determine the molar concentration of acetic acid ([tex]\(HC_2H_3O_2\)[/tex]) in the vinegar, follow these steps:
1. Convert volumes from mL to L:
- Volume of NaOH ([tex]\(V_{\text{NaOH}}\)[/tex]):
[tex]\[
32.45 \text{ mL} = 0.03245 \text{ L}
\][/tex]
- Volume of acetic acid ([tex]\(V_{\text{HC}_2H_3O_2}\)[/tex]):
[tex]\[
10.0 \text{ mL} = 0.01 \text{ L}
\][/tex]
2. Calculate the moles of NaOH used:
The concentration of NaOH ([tex]\(C_{\text{NaOH}}\)[/tex]) is 0.1057 M. The moles of NaOH can be calculated using:
[tex]\[
\text{Moles of NaOH} = V_{\text{NaOH}} \times C_{\text{NaOH}} = 0.03245 \text{ L} \times 0.1057 \text{ M} = 0.003429965 \text{ moles}
\][/tex]
3. Determine the moles of acetic acid (HC_2H_3O_2):
According to the balanced chemical equation, the mole ratio of NaOH to [tex]\(HC_2H_3O_2\)[/tex] is 1:1. Therefore, the moles of [tex]\(HC_2H_3O_2\)[/tex] is equal to the moles of NaOH.
[tex]\[
\text{Moles of } HC_2H_3O_2 = 0.003429965 \text{ moles}
\][/tex]
4. Calculate the concentration of [tex]\(HC_2H_3O_2\)[/tex] in the vinegar:
The molar concentration of [tex]\(HC_2H_3O_2\)[/tex] ([tex]\(C_{\text{HC}_2H_3O_2}\)[/tex]) is found using its moles and volume:
[tex]\[
C_{\text{HC}_2H_3O_2} = \frac{\text{Moles of } HC_2H_3O_2}{V_{\text{HC}_2H_3O_2}} = \frac{0.003429965 \text{ moles}}{0.01 \text{ L}} = 0.3429965 \text{ M}
\][/tex]
Thus, the molar concentration of acetic acid in the vinegar is:
[tex]\[
0.3429965 \text{ M}
\][/tex]
