assuming the density of a 5% acetic acid solution is 1.0g/ml, determine the volume of the acetic acid solution necessary to neutralize 25.0 ml of 0.10 m NaOH. also record this calculation on your report sheet.

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  1. 5% means 5.00g acetic acid in 100 mls of solution. That is equivalent to 50.0 g acid/L.
    The molar mass of acetic acid is about 60.05g/mole
    (50.0 g acid/L.)(1 mol / 60.05g) = 0.8326 mol/L
    The reaction is:
    HC2H3O2 + NaOH --> NaC2H3O2 + H2O
    based on the reaction,
    moles of HC2H3O2 = moles of NaOH
    (0.025 L)(0.10 mol/L) = 0.0025 moles NaOH
    moles of acid = 0.0025 moles HC2H3O2
    V*M = 0.0025 moles
    Substitute to molarity of the acid and solve for V to get the volume in liters. You may want to convert the volume in liters to milliliters at the end.

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  2. 3.001lL

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