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?

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%

the ratio of the conjugate base to the acid required to attain the desired ph.

it is stable as it is in group 0 which have full outer shells of electrons