Dominic made the table below to organize his notes about mixtures. A 1-column table. The first column labeled properties of mixt
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First we need to calculate the number of moles of NaOH titrated.
molar concentration = number of moles / solution volume (liter)
number of moles = molar concentration × solution volume
number of moles of NaOH = 1.5 × 0.0075 = 0.01125 moles
Then we look at the chemical reaction:
CH-COOH + NaOH = CH
COONa + H
O
We can see that 1 mole of acetic acid is reacting with one mole of sodium hydroxide. Then we can conclude that 0.01125 moles of sodium hydroxide reacts with 0.01125 moles of acetic acid.
Now we can get the mass of acetic acid:
number of moles = mass (grams) / molecular mass (g/mol)
mass = number of moles × molecular mass
mass of acetic acid = 0,01125 × 60 = 0.675 g
We assume that the density of the vinegar = 1 g/mL, so the mass percent of acetic acid is:
concentration of acetic acid = (mass of acetic acid / mass of vinegar) × 100
concentration of acetic acid = (0.674 / 7) × 100 = 9.6 %