Question

Perform the following for Part C of this lab:

Answer 1

Answer:

a. 0.0110 L

b. 0.0020 L

c. 0.011 mol

d. 5.5 M

e. 0.66 g

f. 33%

Explanation:

There is some info missing. I will use some values to show you the procedure and then you can replace them with your values.

Titrant (NaOH) concentration: 1.0 M

Vinegar volume: 2.0 mL

Initial buret reading (initial NaOH volume): 0.1 mL

Final buret reading (final NaOH volume): 11.1 mL

a. Calculate the volume of NaOH that was added to the vinegar. Convert this volume to liters. Show your work.

The volume of NaOH is the difference between the final and the initial buret reading.

11.1 mL - 0.1 mL = 11.0 mL × (1 L/1000 mL) = 0.0110 L

b. Convert the measured volume of vinegar to liters. Show your work.

2.0 mL × (1 L/1000 mL) = 0.0020 L

c. Calculate the moles of NaOH using the volume and molarity of NaOH. Show your work. moles = molarity x volume

moles = molarity × volume

moles = (1.0 mol/L) × 0.0110 L = 0.011 mol

d. Since the reaction ratio is 1:1, the moles of acetic acid in the vinegar is equal to the moles of NaOH reacted during the titration. Calculate the molarity of the acetic acid in the vinegar. Show your work. molarity = moles / volume

molarity = moles / volume

molarity = 0.011 mol/0.0020 L = 5.5 M

e. Calculate the grams of acetic acid in the vinegar. Show your work. mass = moles x molar mass (g/mol)

mass = moles × molar mass

mass = 0.011 mol × 60.05 g/mol = 0.66 g

f. Assuming that the density of vinegar is very close to 1.0 g/mL, the 2.0 mL sample of vinegar used in the titration should weigh 2.0  g. Use this to calculate the mass % of acetic acid in the vinegar sample. mass % = (mass acetic acid / mass vinegar) * 100%

mass % = (mass acetic acid / mass vinegar) * 100%

mass % = (0.66 g /2.0 g) * 100% = 33%