Respuesta :
The question requires us to calculate the molar concentration of an acetic acid solution, given that it was necessary 36.4 mL of a 0.100 M NaOH solution to neutralize 50.0 mL of the acid.
The following information was provided by the question:
concentration of NaOH solution = C(OH-) = 0.100 M
volume of NaOH solution = V(OH-) = 36.4 mL
volume of CH3COOH solution = V(H+) = 50.0 mL
Acetic acid (CH3COOH) and sodium hydroxide (NaOH) react according to the following reaction:
[tex]\text{CH}_3COOH_{}+NaOH_{}\to CH_3COONa+H_2O[/tex]Considering that at the equivalent point the number of moles of acid (H+) is the same as the number of moles of base (OH-) and taking the stoichiometry of the reaction into consideratio (1 : 1), we can write:
[tex]At\text{ equivalent point}\to n_{H^+}n_{OH^-}\to n_{CH_3COOH}=n_{NaOH}[/tex]We can also write the number of moles of a compound (n) in terms of its molar concentration and volume:
[tex]\begin{gathered} \text{molar concentration = }\frac{number\text{ of moles}}{\text{volume}}\to C=\frac{n}{V} \\ \\ n=C\times V \end{gathered}[/tex]Thus, considering the equivalent point, we can say:
[tex]n_{CH_3COOH}=n_{NaOH}\to C_{CH_3COOH}\times V_{CH_3COOH}=C_{NaOH}\times V_{NaOH}[/tex]Now, we can apply the values of volume and concentration provided by the question to the equation above and obtain the concentration of the acid:
[tex]\begin{gathered} C_{CH_3COOH}\times V_{CH_3COOH}=C_{NaOH}\times V_{NaOH} \\ C_{CH_3COOH}\times(50.0mL)=(0.100M)\times(36.4mL) \\ C_{CH_3COOH}=\frac{(0.100M)\times(36.4mL)}{(50.0mL)}=0.0728M \end{gathered}[/tex]Therefore, the molar concentration of the acetic acid solution is 0.0728 M and the best option to answer this question is the third one.
