Textbook Question
Calculate the volume, in milliliters, of a 0.150 M NaOH solution that will completely neutralize each of the following:
a. 25.0 mL of a 0.288 M HCl solution
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10. Acids and Bases
Strong Acid Strong Base Titrations (Simplified)
Multiple Choice
Vinegar is a solution of acetic acid, CH3COOH, dissolved in water. A 5.54 g sample of vinegar was neutralized by 30.10 mL of 0.100 M NaOH. What is the mass percent of acetic acid in the vinegar?
____CH3COOH (aq) + ____NaOH (aq) → ____CH3COONa (aq) + ____H2O (l)
A
2.53 %
B
3.12 %
C
1.63 %
D
3.26 %
Verified step by step guidance1
Write the balanced chemical equation for the neutralization reaction between acetic acid and sodium hydroxide: \(\mathrm{CH_3COOH (aq) + NaOH (aq) \rightarrow CH_3COONa (aq) + H_2O (l)}\). This shows a 1:1 mole ratio between acetic acid and NaOH.
Calculate the moles of NaOH used in the reaction using the formula: \(\text{moles NaOH} = M \times V\), where \(M\) is the molarity (0.100 M) and \(V\) is the volume in liters (30.10 mL = 0.03010 L).
Since the mole ratio of acetic acid to NaOH is 1:1, the moles of acetic acid neutralized are equal to the moles of NaOH calculated.
Calculate the mass of acetic acid in the vinegar sample by multiplying the moles of acetic acid by its molar mass (approximately 60.05 g/mol): \(\text{mass of } \mathrm{CH_3COOH} = \text{moles} \times 60.05 \ \mathrm{g/mol}\).
Determine the mass percent of acetic acid in the vinegar using the formula: \(\text{mass \%} = \left( \frac{\text{mass of acetic acid}}{\text{mass of vinegar sample}} \right) \times 100\%\), where the mass of the vinegar sample is 5.54 g.
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