Determine the molarity of acetic acid if 13.45 mL of acetic acid requires 22.73 mL of 0.112 M NaOH for complete neutralization. Assume the reaction is 1:1.
A
0.112 M
B
0.067 M
C
0.189 M
D
0.224 M
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1
Identify the balanced chemical equation for the neutralization reaction: \( \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \). This shows a 1:1 molar ratio between acetic acid and sodium hydroxide.
Calculate the moles of NaOH used in the reaction. Use the formula: \( \text{moles of NaOH} = \text{Molarity of NaOH} \times \text{Volume of NaOH in liters} \). Convert 22.73 mL to liters by dividing by 1000.
Since the reaction is 1:1, the moles of acetic acid will be equal to the moles of NaOH calculated in the previous step.
Calculate the molarity of acetic acid using the formula: \( \text{Molarity of acetic acid} = \frac{\text{moles of acetic acid}}{\text{Volume of acetic acid in liters}} \). Convert 13.45 mL to liters by dividing by 1000.
Substitute the values obtained into the molarity formula to find the molarity of acetic acid.