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Suppose you are titrating vinegar, which is an acetic acid solution of unknown concentration, with a sodium hydroxide solution according to the equation
[tex]\[ HC_2H_3O_2 + NaOH \longrightarrow H_2O + NaC_2H_3O_2 \][/tex]

If you require 32.45 mL of 0.1057 M NaOH solution to titrate 10.0 mL of [tex]\( HC_2H_3O_2 \)[/tex] solution, what is the molar concentration of acetic acid in the vinegar?

Type answer:
[tex]\[ \square \][/tex]

Sagot :

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]