Analysis of a Commercial Bleach Lab AP Chem.
Analysis of a Commercial Bleach
I. Purpose: The purpose of the lab is to analyze commercial bleach. Commercial bleach is diluted, iodide is added, and the iodine that forms is titrated with a standard solution of sodium thiosulfate, to determine the concentration of NaClO in the commercial bleach.
II. Pre-Lab Questions:
1. What is meant by a “titration”? A titration is the process of reacting a solution of unknown concentration with one of known concentration.
2. A solution of household vinegar (a mixture of acetic acid and water) is to be analyzed. A pipet is used to measure out 10.0 mL of the vinegar, which is placed in a 250-mL volumetric flask. Distilled water is added until the total volume of the solution is 250 mL. A 25.0-mL portion of the diluted solution is measured out with a pipet and titrated with a standard solution of sodium hydroxide.
The neutralization reaction is as follows:
HC2H3O2 (aq) + OH– (aq) à C2H3O2– (aq) + H2O (l)
It is found that 16.7 mL of 0.0500 M NaOH is needed to titrate 25.0 mL of the diluted vinegar. Calculate the molarity of the diluted vinegar.
(0.0167 L NaOH) x (0.0500 moles NaOH / 1 L NaOH) = 8.35 x 10 –4 moles NaOH
1:1 mole ratio for sodium hydroxide and acetic acid; 8.35 x 10 –4 moles acetic acid
(8.35 x 10 –4 moles acetic acid) / (0.025 L solution) = 0.0334 M dilute vinegar
3. Calculate the molarity of the household vinegar.
M1V1 = M2V2; (X M concentrated)(10 mL) = (0.0334 M dilute)(250 mL)
0.835 M household vinegar
4. The household vinegar has a density of 1.05 g/mL. Calculate the percent by mass of acetic acid in the household vinegar.
(1 L household vinegar)(1.05 g/mL) = 1050 g household vinegar
(1 L household vinegar)(0.835 mol acetic acid / L solution) = 0.835 moles acetic acid
(0.835 moles acetic acid) x (60.052 g acetic acid / mol acetic acid) = 50.14 g acetic acid