LAB PH
PH at the UN
Alfredo Vidal Ceballos
7/10/2014
Group Support Part 1:
Solution Group Name and M. cons. Chem formula Type Experimental pH
1 Deidre Acetic Acid 0.1M HC2H3O2 Weak Acid 3.03 3.03
2 Acetic Acid 1M HC2H3O2 Weak Acid 2.59 2.54
3 Hydrochloric Acid 0.5M HClStrong Base 0.85 0.90
4 Jonah Nitric Acid 0.1M HNO3 Strong Acid 2.14 2.14
5 Sodium Acetate 1M NaC2H3O2 Weak Base 9.42 9.42
6 Potassium Dihydrogen Phosphate 0.1M KH2PO4 Weak Acid 5.47 5.47
7 Andy Potassium Hydrogen Phosphate 0.05M K2HPO4 Weak Base 7.71 7.73
8 Ammonium Chloride 0.2M NH4Cl Weak Acid 4.65 4.67
9 Sodium Hydroxide 0.1M NaOHStrong Base 11.95 11.95
10 Rachel Sodium hydroxide 0.01M NaOHBase 11.61 11.72
11 Ammonia 0.1M NH3 Base 9.53 9.48
12 Vinegar 0.83M HC2H3O2 Acid 2.83 2.94
13 Petr Sodium 0.5M NaC6H5O7 Weak Base 9.23 9.21
14 Sodium Sulfate 0.25M Na2SO4 Weak Base 7.24 7.23
15 Potassium Sulfate 0.25M K2SO4 Weak Base 7.37 7.58
Dilution of 1M of acetic acid to 0.1M and 0.5M
0.5M of HCl dilution: Vi(0.5M)(0.025L)1Mx1000=12.5mLof HCl used + 12.5mL of water.
0.1M of HC2H3O2 dilution: Vi(0.1M)(0.025mL)1Mx1000=2.5mL HC2H3O2 used + 22.5mL of water.
HC2H3O2 was provided in a concentration of 1M.
Solution Observations:
All the solutions where clear liquids before and after they were diluted.
Compound name and concentration Calculations Theoretical pH
Acetic Acid 0.1M HC2H3O2 + H2O CH3COO- + H3O+
0.1M – x x xKa=1.8x10-5 = x20.1M= (1.8x10-5 x 0.1M) = 1.3416x10-3
= -log(1.3416x10-3) = 2.87 2.87
Acetic Acid 1M (1.8x10-5x 1M)= 4.24x10-3
= -log (4.24x10-3) = 2.37 2.37
Hydrochloric Acid 0.5M pH = -log (0.5) = 0.30 0.30
Nitric Acid 0.1M pH = -log(0.1) = 1 1
Sodium Acetate 1M Kb=1.0x10-141.8x10-5=5.55x10-10pOH = -log(5.55x10-10) = 4.627
14 – 4.627 = 9.37 9.37
Potassium Dihydrogen Phosphate 0.1M (6.2x10-8)0.1M = 7.87x10-5 pH = -log(7.87x10-5) = 4.10 4.10
Potassium Hydrogen Phosphate 0.05M (4.8x10-13)(0.05) = 1.54x10-7 pH = -log(1.54x10-7) = 6.80 6.80
Ammonium Chloride 0.2M
Alfredo Vidal Ceballos
7/10/2014
Group Support Part 1:
Solution Group Name and M. cons. Chem formula Type Experimental pH
1 Deidre Acetic Acid 0.1M HC2H3O2 Weak Acid 3.03 3.03
2 Acetic Acid 1M HC2H3O2 Weak Acid 2.59 2.54
3 Hydrochloric Acid 0.5M HClStrong Base 0.85 0.90
4 Jonah Nitric Acid 0.1M HNO3 Strong Acid 2.14 2.14
5 Sodium Acetate 1M NaC2H3O2 Weak Base 9.42 9.42
6 Potassium Dihydrogen Phosphate 0.1M KH2PO4 Weak Acid 5.47 5.47
7 Andy Potassium Hydrogen Phosphate 0.05M K2HPO4 Weak Base 7.71 7.73
8 Ammonium Chloride 0.2M NH4Cl Weak Acid 4.65 4.67
9 Sodium Hydroxide 0.1M NaOHStrong Base 11.95 11.95
10 Rachel Sodium hydroxide 0.01M NaOHBase 11.61 11.72
11 Ammonia 0.1M NH3 Base 9.53 9.48
12 Vinegar 0.83M HC2H3O2 Acid 2.83 2.94
13 Petr Sodium 0.5M NaC6H5O7 Weak Base 9.23 9.21
14 Sodium Sulfate 0.25M Na2SO4 Weak Base 7.24 7.23
15 Potassium Sulfate 0.25M K2SO4 Weak Base 7.37 7.58
Dilution of 1M of acetic acid to 0.1M and 0.5M
0.5M of HCl dilution: Vi(0.5M)(0.025L)1Mx1000=12.5mLof HCl used + 12.5mL of water.
0.1M of HC2H3O2 dilution: Vi(0.1M)(0.025mL)1Mx1000=2.5mL HC2H3O2 used + 22.5mL of water.
HC2H3O2 was provided in a concentration of 1M.
Solution Observations:
All the solutions where clear liquids before and after they were diluted.
Compound name and concentration Calculations Theoretical pH
Acetic Acid 0.1M HC2H3O2 + H2O CH3COO- + H3O+
0.1M – x x xKa=1.8x10-5 = x20.1M= (1.8x10-5 x 0.1M) = 1.3416x10-3
= -log(1.3416x10-3) = 2.87 2.87
Acetic Acid 1M (1.8x10-5x 1M)= 4.24x10-3
= -log (4.24x10-3) = 2.37 2.37
Hydrochloric Acid 0.5M pH = -log (0.5) = 0.30 0.30
Nitric Acid 0.1M pH = -log(0.1) = 1 1
Sodium Acetate 1M Kb=1.0x10-141.8x10-5=5.55x10-10pOH = -log(5.55x10-10) = 4.627
14 – 4.627 = 9.37 9.37
Potassium Dihydrogen Phosphate 0.1M (6.2x10-8)0.1M = 7.87x10-5 pH = -log(7.87x10-5) = 4.10 4.10
Potassium Hydrogen Phosphate 0.05M (4.8x10-13)(0.05) = 1.54x10-7 pH = -log(1.54x10-7) = 6.80 6.80
Ammonium Chloride 0.2M