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22.04.2020 •
Chemistry
You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.1950 M NaOH solution to exactly neutralize this sample (phenolphthalein was used as an indicator). What is the molarity of the acetic acid solution? What is the percentage of acetic acid in the solution?
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Ответ:
Concentration acetic acid = 0.27885 M
% acetic acid = 0.69%
Explanation:
You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.1950 M NaOH solution to exactly neutralize this sample (phenolphthalein was used as an indicator). What is the molarity of the acetic acid solution?
what is the percentage of acetic acid in the solution? Assume the density of the solution is 1 g/ml.
Step 1: Data given
Volume of acetic acid = 25.00 mL = 0.025 L
Volume of NaOH = 35.75 mL = 0.03575 L
Molarity of NaOH = 0.1950 M
Step 2: The balanced equation
CH3COOH + NaOH → CH3COONa + H2O
Step 3: Calculate moles
Moles = molarity * volume
Moles NaOH = 0.1950 M * 0.03575 L
Moles NaOH = 0.00697125 moles
Step 4: Calculate concentration of acetic acid
We need 0.00697125 moles of acetic acid to neutralize NaOH
Concentration = moles / volume
Concentration = 0.00697125 moles / 0.025 L
Concentration = 0.27885 M
Step 5: Calculate mass of acetic acid
Mass acetic acid = moles * molar mass
Mass acetic acid = 0.00697125 moles * 60.05g/mol
Mass acetic acid = 0.4186 grams
Step 6: Calculate mass of sample
Total volume = 60.75 mL = 0.06075 L
Mass of sample 60.75 mL * 1g/mL = 60.75 grams
Step 7: Calculate the percentage of acetic acid in the solution
% acetic acid = (0.4186 grams / 60.75 grams ) * 100 %
% acetic acid = 0.69%
Ответ: