Question

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?

Answer 1

Concentration acetic acid = ‬0.27885 M

% acetic acid = 0.69%

Step-by-step 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%

Answer 2

0.2788 M

1.674 %(m/V)

Step-by-step explanation:

Step 1: Write the balanced equation

NaOH + CH₃COOH → CH₃COONa + H₂O

Step 2: Calculate the reacting moles of NaOH

`0.03575 L * (0.1950mol)/(L) = 6.971 * 10^(-3) mol`

Step 3: Calculate the reacting moles of CH₃COOH

The molar ratio of NaOH to CH₃COOH is 1:1.

`6.971 * 10^(-3) molNaOH * (1molCH_3COOH)/(1molNaOH) = 6.971 * 10^(-3) molCH_3COOH`

Step 4: Calculate the molarity of the acetic acid solution

`M = (6.971 * 10^(-3) mol)/(0.02500L) =0.2788 M`

Step 5: Calculate the mass of acetic acid

The molar mass of acetic acid is 60.05 g/mol.

`6.971 * 10^(-3) mol * (60.05g)/(mol) =0.4186 g`

Step 6: Calculate the percentage of acetic acid in the solution

`(0.4186g)/(25.00mL) * 100\% = 1.674 \%(m/V)`