Answer:
3.0 mL
Explanation:
Let's consider the neutralization between acetic acid and NaOH.
CH₃COOH + NaOH → CH₃COONa + H₂O
The moles of NaOH are:
25.0 × 10⁻³ L × 0.10 mol/L = 2.5 × 10⁻³ mol
The molar ratio of CH₃COOH to NaOH is 1:1. The moles of CH₃COOH are 2.5 × 10⁻³ mol.
The molar mass of CH₃COOH is 60.05 g/mol. The mass of CH₃COOH is:
2.5 × 10⁻³ mol × (60.05 g/mol) = 0.15 g
The mass percent of acetic acid is 5%. The mass of the solution of acetic acid is:
0.15 g CH₃COOH × (100 g solution/ 5 g CH₃COOH) = 3.0 g solution
The density of the acetic acid solution is 1.0 g/mL. The volume of the solution is:
3.0 g × (1 mL/1.0 g) = 3.0 mL
Volume of the acetic acid solution necessary to neutralize is 3 ml.
What information do we have?
Density of 5% acetic acid solution = 1 g/mL
Amount neutralize 25 ml of 0.10M NaOH
5% acetic acid solution = 5 g acetic acid in 100 ml solution
Molarity of acetic acid solution = [5/60] × 0.1
Molarity of acetic acid solution = [0.083]0.1
Molarity of acetic acid solution = 0.83 M
Moles NaOH = 0.1 M × 25 ml
Moles NaOH= 2.5 mol
Volume acetic acid needed = 2.5/0.83
Volume acetic acid needed = 3.0 ml
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