Question

The distinctive odor of vinegar is due to acetic acid, HC₂H₃O₂, which reacts with sodium hydroxide. If a 3.45 mL sample of vinegar needs 42.5 mL of 0.115 M NaOH to be neutralized, how many grams of acetic acid are in a 1.00 L bottle of this vinegar?​

Answer 1

To find the number of grams of acetic acid in a 1.00 L bottle of vinegar, we can use the volume of sodium hydroxide used in the neutralization reaction. By knowing the ratio of acetic acid to sodium hydroxide, we can calculate the moles of acetic acid and then convert it to grams using the molar mass.

Step-by-step explanation:

To calculate the number of grams of acetic acid in a 1.00 L bottle of vinegar, we need to use the given information about the neutralization reaction between acetic acid and sodium hydroxide. From the reaction equation:

2CH3 CO₂ H(aq) + NaOH(aq) → 2CH3 CO₂ Na(aq) + H2O(l)

We can see that the ratio of acetic acid to sodium hydroxide is 2:1. Therefore, the number of moles of acetic acid can be determined by multiplying the volume of NaOH (42.5 mL) by its molarity (0.115 M), and then dividing by 2. To convert moles to grams, we need to use the molar mass of acetic acid. The molar mass of acetic acid is 60.05 g/mol. Finally, we can calculate the number of grams of acetic acid in a 1.00 L bottle of vinegar by using the equation:

grams of acetic acid = (moles of acetic acid) x (molar mass of acetic acid)

Answer 2

Step-by-step explanation:

CH₃COOH + NaOH = CH₃COONa + H₂O .

42.5 mL of .115 M of NaOH will contain .0425 x .115 moles of NaOH

= 48.875 x 10⁻⁴ moles NaOH

It will react with same number of moles of acetic acid

So number of moles of acetic acid in 3.45 mL = 48.875 x 10⁻⁴

number of moles of acetic acid in 1000 mL = 48.875 x 10⁻⁴ x 10³ / 3.45 moles

= 1.4167 moles

= 1.4167 x 60 gram

= 85 grams .

So 85 grams of acetic acid will be contained in one litre of acetic acid.