3.30 mL of vinegar needs 43.0 mL of 0.150 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.70 qt sample of this vinegar?

189g

I agree.

To solve this problem, we need to use stoichiometry to relate the volume of NaOH used to the amount of acetic acid present in the vinegar.

First, let's convert the given volume of vinegar from quarts to milliliters since the volume of NaOH is given in milliliters.

1 quart (qt) is equal to 946.353 milliliters (mL). Therefore, 1.70 quarts equals:

1.70 qt * 946.353 mL/qt = 1,607.9 mL

Next, let's set up the stoichiometric relationship between NaOH and acetic acid:

1 mole of NaOH reacts with 1 mole of acetic acid, according to the balanced equation:

CH3COOH + NaOH → CH3COONa + H2O

The balanced equation shows a 1:1 mole ratio between NaOH and acetic acid.

Since the concentration of NaOH is given as 0.150 M, we can calculate the number of moles of NaOH used in the titration:

Molarity (M) = moles/volume (L)

0.150 M = moles/0.043 L

moles of NaOH = 0.150 M * 0.043 L = 0.00645 moles

Since the stoichiometry is 1:1, this means that 0.00645 moles of acetic acid are present in the vinegar.

Finally, let's calculate the mass of acetic acid using the molar mass of acetic acid (CH3COOH), which is approximately 60.052 grams/mole:

mass (g) = moles * molar mass (g/mole)

mass of acetic acid = 0.00645 moles * 60.052 g/mole = 0.387 grams

Therefore, there are approximately 0.387 grams of acetic acid in a 1.70 quart sample of this vinegar.