Average Dilution Ratio of Ammonia in Lake Michigan:
1 mL of ammonia: 40 mL of alkaline water pH of Windex Solution at this ratio: 9 Volume of Windex: 10 mL Weight Percentage Ammonia in Windex: 5% Mass of Ammonia: 0.5 grams NH3 Molar Mass of Ammonia: 17.04 g/mol NH3 Moles of Ammonia: 0.029 mol NH3 Volume of Solution: 410 mL (0.410 L) Molarity of Solution: M = n/v M = 0.029 / 0.410 M = 0.071 M |
Least Influential Dilution Ratio of Ammonia in Lake Michigan: 1 mL of ammonia: 70 mL of alkaline water
Constant pH of Windex Solution at this ratio: 8 Volume of Windex: 10 mL Weight Percentage Ammonia in Windex: 5% Mass of Ammonia: 0.5 grams NH3 Molar Mass of Ammonia: 17.04 g/mol NH3 Moles of Ammonia: 0.029 mol NH3 Volume of Solution: 710 mL (0.710 L) Molarity of Solution: M = n/v M = 0.029 / 0.710 M = 0.041 M |
Check Previous Calculations:
M1V1 = M2V2
(0.071)(410 mL) = M (710 mL) M = 0.041 M Percent Error: (0.041 - 0.041) / 0.041 x 100 = 0.0 % Error |
M1V1 = M2V2
M(410 mL) = (0.041)(710 mL) M = 0.071 M Percent Error: (0.071 - 0.071) / 0.071 x 100 = 0.0% Error |
pH Level of Ammonia in Windex at 1:40 Dilution
[pOH] = -log(0.072) [pOH] = 1.1 14 - 1.1 = 12.9 pH of Ammonia in Windex = 12.9 |
pH Level of Ammonia in Windex at 1:70 Dilution
[pOH] = -log(0.041) [pOH] = 1.4 14 - 1.4 = 12.6 pH of Ammonia in Windex = 12.6 |
Experimental Ratio of Vinegar in Lake Michigan:
Amount of Vinegar: 15 mL
Amount of Alkaline Water: 400 mL + an additional 15 mL
Average Dilution Ratio of Vinegar in Lake Michigan:
3 mL of vinegar : 83 mL of alkaline water
pH of Vinegar Solution at this ratio: 7
Density of Acetic Acid: 1.05 g/mL
Volume of Vinegar: 15 mL
Weight Percentage of Acetic Acid (Average of Titration Trials): 4.8%
Mass of Acetic Acid: 0.72 grams CH3COOH
Molar Mass of Acetic Acid: 60.06 g/mol CH3COOH
Moles of Acetic Acid: 0.012 mol CH3COOH
Volume of Solution: 430 mL (0.430 L)
Molarity of Solution: M = n/v
M = 0.012 / 0.430
M = 0.028 M
Amount of Vinegar: 15 mL
Amount of Alkaline Water: 400 mL + an additional 15 mL
Average Dilution Ratio of Vinegar in Lake Michigan:
3 mL of vinegar : 83 mL of alkaline water
pH of Vinegar Solution at this ratio: 7
Density of Acetic Acid: 1.05 g/mL
Volume of Vinegar: 15 mL
Weight Percentage of Acetic Acid (Average of Titration Trials): 4.8%
Mass of Acetic Acid: 0.72 grams CH3COOH
Molar Mass of Acetic Acid: 60.06 g/mol CH3COOH
Moles of Acetic Acid: 0.012 mol CH3COOH
Volume of Solution: 430 mL (0.430 L)
Molarity of Solution: M = n/v
M = 0.012 / 0.430
M = 0.028 M
Titration Experiment Acetic Acid Calculations:
Trial #1:
Acetic Acid Molarity: M = 0.40 M (20.1 mL) / 10 mL M = 0.80 M CH3COOH Trail #2: Acetic Acid Molarity: M = 0.40 M (20.5 mL) / 10 mL M= 0.82 M CH3COOH Trail #3: Acetic Acid Molarity: M = 0.40 M (19.7 mL) / 10 mL M = 0.79 M CH3COOH Trail #4: Acetic Acid Molarity: M = 0.40 M (19.9 mL) / 10 mL M= 0.80 M CH3COOH Trail #5: Acetic Acid Molarity: M = 0.40 M (20.2 mL) / 10 mL M= 0.81 M CH3COOH |
Mass of Acetic Acid / 1 Liter of Vinegar: Mass = 0.80 M (60.06 g/mol) Mass = 48 grams CH3COOH Mass of Acetic Acid / 1 Liter of Vinegar: Mass = 0.82 M (60.06 g/mol) Mass = 49 grams CH3COOH Mass of Acetic Acid / 1 Liter of Vinegar: Mass = 0.79 M (60.06 g/mol) Mass = 47 grams CH3COOH Mass of Acetic Acid / 1 Liter of Vinegar: Mass = 0.80 M (60.06 g/mol) Mass = 48 grams CH3COOH Mass of Acetic Acid / 1 Liter of Vinegar: Mass = 0.81 M (60.06 g/mol) Mass = 49 grams CH3COOH |
Weight % of Acetic Acid in Vinegar: w% = (48/1) / (1000/1) x 100 w% = 4.8 % CH3COOH Weight % of Acetic Acid in Vinegar: w% = (49/1) / (1000/1) x 100 w% = 4.9 % CH3COOH Weight % of Acetic Acid in Vinegar: w% = (47/1) / (1000/1) x 100 w% = 4.7 % CH3COOH Weight % of Acetic Acid in Vinegar: w% = (48/1) / (1000/1) x 100 w% = 4.8 % CH3COOH Weight % of Acetic Acid in Vinegar: w% = (49/1) / (1000/1) x 100 w% = 4.9 % CH3COOH |