Diluted white vinegar contains ethanoic acid (CH3COOH\text{CH}_3\text{COOH}CH3COOH). The concentration of ethanoic acid can be determined by titration with aqueous sodium hydroxide, NaOH(aq)\text{NaOH}(aq)NaOH(aq).
A student titrated a 25.0 cm325.0\text{ cm}^325.0 cm3 sample of diluted white vinegar, which reacted with exactly 18.50 cm318.50\text{ cm}^318.50 cm3 of 0.120 mol/dm30.120\text{ mol/dm}^30.120 mol/dm3 aqueous sodium hydroxide.
The equation for the reaction is:
CH3COOH(aq)+NaOH(aq)→CH3COONa(aq)+H2O(l) \text{CH}_3\text{COOH}(aq) + \text{NaOH}(aq) \rightarrow \text{CH}_3\text{COONa}(aq) + \text{H}_2\text{O}(l) CH3COOH(aq)+NaOH(aq)→CH3COONa(aq)+H2O(l)Calculate the amount, in moles, of sodium hydroxide in 18.50 cm318.50\text{ cm}^318.50 cm3 of a 0.120 mol/dm30.120\text{ mol/dm}^30.120 mol/dm3 solution.
Deduce the amount, in moles, of ethanoic acid in the 25.0 cm325.0\text{ cm}^325.0 cm3 sample of diluted white vinegar.
Calculate the concentration, in mol/dm3\text{mol/dm}^3mol/dm3, of ethanoic acid in the diluted white vinegar.
Calculate the concentration, in g/dm3\text{g/dm}^3g/dm3, of ethanoic acid in the diluted white vinegar. (Relative atomic masses: H=1.0\text{H} = 1.0H=1.0, C=12.0\text{C} = 12.0C=12.0, O=16.0\text{O} = 16.0O=16.0)
To be classified as "standard strength", a commercial vinegar must contain at least 5.00 g/dm35.00\text{ g/dm}^35.00 g/dm3 of ethanoic acid.
Use the value you have calculated in (d) to state whether this vinegar can be classified as "standard strength".
