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15 January, 17:31

Assuming the density of a 5%acetic acid solution is

1.0g/mL, determine the volume of the aceticacid solution necessary

to neutralize 25.0mL of 0.10M NaOH.

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Answers (1)
  1. 15 January, 18:49
    0
    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
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