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If 36.3 mL of 0.152 M NaOH is required to neutralize 25.00 mL of an HC₂H₃O₂ solution, what is the molarity of the acetic acid?

A) 0.152 M
B) 0.100 M
C) 0.090 M
D) 0.080 M

Answer :

Final answer:

Calculating the molarity of acetic acid involves using the volume and molarity of NaOH to find the moles of NaOH, which equates to the moles of acetic acid due to the 1:1 molar ratio. By dividing the moles of acetic acid by the volume of the acetic acid solution, one obtains the molarity. The result was different from the options provided, indicating a potential typo.

Explanation:

Calculating the Molarity of Acetic Acid

The question asks for the molarity of an acetic acid solution after a titration with NaOH. The reaction between NaOH and HC2H3O2 is a neutralization reaction in which NaOH and acetic acid react in a 1:1 molar ratio:

NaOH + HC2H3O2 → NaC2H3O2 + H2O

Since 36.3 mL of 0.152 M NaOH is used to completely neutralize the acetic acid, we can calculate the moles of NaOH used:

Moles of NaOH = 0.0363 L * 0.152 mol/L = 0.005516 moles

Given the 1:1 molar ratio, the moles of acetic acid will also be 0.005516 moles. To find the molarity of acetic acid, divide the moles of acetic acid by the volume of the acetic acid solution (in liters):

Molarity of HC2H3O2 = Moles of HC2H3O2 / 0.02500 L = 0.005516 moles / 0.02500 L = 0.22064 M

However, this result is not one of the answer options provided, which indicates that there might be a typo in the question or the potential answers. A correct procedure was followed to obtain a molarity, but further clarification is needed.

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