Purpose: To find the percent composition of sodium hypochlorite in commercial bleach.
| |Trial 1 |Trial 2 |Trial 3 |
|Initial buret reading |3.7 |6.1 |4.3 |
|Final buret reading |24.9 |27.3 |25.5 |
|Volume Used |21.2 mL |21.2 mL |21.2 mL |
1. What is meant by a “titration”?
When you take a solution with a known concentration and known volume and react it with a known volume and unknown concentration of another solution to find its concentration.
2. A solution of household vinegar (a mixture of acetic acid and water) is to be analyzed. A pipette 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 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.
HC2H3O2 (aq) + OH-1 (aq) -> 2H3O2-1 (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.
10.67 mL x .05 M = 25. mL x ?M
?M = .0334M
3. Calculate the molarity of the household vinegar.
.0334 x 250/10 = .835 M
4. The household vinegar has a density of 1.05 g/mL. Calculate the percent by mass of acetic acid in the household vinegar.
1.05 g/mL x 1000 mL = 1050 g of vinegar per liter
.835 x 60.0 g/mole = 50.1 g
50.1 g CH3COOH/1050 g Vinegar = 4.77 %
Data and Calculations
1. Two moles of sodium thiosulfate are equivalent to one mole...