Question 5.9: Using Titration Data to Establish the Concentrations of Acid......
Using Titration Data to Establish the Concentrations of Acids and Bases
Vinegar is a dilute aqueous solution of acetic acid produced by the bacterial fermentation of apple cider, wine, or other carbohydrate material. The legal minimum acetic acid content of vinegar is 4% by mass. A 5.00 mL sample of a particular vinegar is titrated with 38.08 mL of 0.1000 M NaOH. Does this sample exceed the minimum limit? (Vinegar has a density of about 1.01 g/mL.)
Analyze
Acetic acid, CH_{3}COOH, is a weak acid and NaOH is a strong base. The reaction between CH_{3}COOH and NaOH is an acid–base neutralization reaction. We start by writing a balanced chemical equation for the reaction. We must convert mL NaOH to g CH_{3}COOH. The necessary conversions are as follows:
mL NaOH → L NaOH → mol NaOH → mol CH_{3}COOH → g CH_{3}COOH
Solve
The balanced chemical equation for the reaction is given below.
CH_{3}COOH(aq) + NaOH(aq) → NaCH_{3}COO(aq) + H_{2}O(l)
? g CH_{3}COOH = 38.08 mL × \frac{1 L}{1000 mL} × \frac{0.1000 mol NaOH}{1 L} × \frac{1 mol CH_{3}COOH}{1 mol NaOH} × \frac{60.05 g CH_{3}COOH}{1 mol CH_{3}COOH}
= 0.2287 g CH_{3}COOH
This mass of CH_{3}COOH is found in 5.00 mL of vinegar of density 1.01 g/mL. The percent mass of CH_{3}COOH is
% CH_{3}COOH = \frac{0.2287 g CH_{3}COOH}{5.00 mL vinegar} × \frac{1 mL vinegar}{1.01 g vinegar} × 100% = 4.53% CH_{3}COOH
The vinegar sample exceeds the legal minimum limit but only slightly. There is also a standard for the maximum amount of acetic acid allowed in vinegar. A vinegar producer might use this titration technique to ensure that the vinegar stays between these limits.
Assess
Such problems as this one involve many steps or conversions. Try to break the problem into simpler ones involving fewer steps or conversions. It may also help to remember that solving a stoichiometry problem involves three steps: (1) converting to moles, (2) converting between moles, and (3) converting from moles. Use molarities and molar masses to carry out volume–mole conversions and gram–mole conversions, respectively, and stoichiometric factors to carry out mole–mole conversions. The stoichiometric factors are constructed from a balanced chemical equation.