Standardizing a Solution of Potassium Hydroxide
Jay Patel
10/21/11-10/21/11
Lab Partners: Isha Dihora, Shivandu Patel, Dilan Kapadia
Standardizing a Solution of Potassium Hydroxide
Abstract: In this lab, a prepared solution of Potassium Hydroxide will be standardized. The solution we will be standardized by performing multiple calculations to ensure the upmost accuracy. The acid used for this titration will be KHP (C4H5KO4). Phenolphthalein we be added to the beaker of the dissolved acid before the titration will be added. The titration will stop when the solution turns pink and the color doesn’t go away.
Procedure: The procedure is stated in my notebook
Data/Observations:
Observations of all trials | * The solid KHP dissolved in distilled water * The Solution turned stayed pink after enough KOH was added to the dissolved KHP. |
Trial 1: 1.5316g KHP
50.000 mL KOH in burette
28.000 mL KOH in beaker Trial 2: 1.5408g KHP
50.000mL KOH in burette
31.500mL KOH in beaker
Trial 3: 1.5380g KHP
50.000mL KOH in burette
31.500mL KOH in beaker
Analysis:
0.025L × .300M1L= .0075mol KOH
KHP (C8H5KO4) molar mass: (12.0101 ×8) + (1.00794×5) + (39.0983) + (15.9994×4) = 204.2156g
.0075mol KOH ×1mol KHP1mol KOH ×204.2156g KHP1mol KHP=1.5316g KHP for solid
1.5316g×1mole204.2156g=7.500×10-3moles 7.500×10-3moles.02800L=0.26787M
1.5408g×1mole204.2156g=7.545×10-3moles 7.545×10-3moles.03150L=0.23954M
1.538g×1mole204.2156g=7.532×10-3 7.532×10-3.03150L=0.23910M
.26787+.23954+.23910 =.74651 .746513=0.24884M KOH Trial 1: 0.26787 M Trial 2: 0.23954 M Trial 3: 0.23910 M
Accepted value: 0.24884 M Percent yield: .24884.2583×100%=96.33% Percent error:.24884-.2583.2583×100%=3.662%
Conclusion: In this experiment, a solution of KOH was standardized with titration. The KOH reacted with KHP which was
10/21/11-10/21/11
Lab Partners: Isha Dihora, Shivandu Patel, Dilan Kapadia
Standardizing a Solution of Potassium Hydroxide
Abstract: In this lab, a prepared solution of Potassium Hydroxide will be standardized. The solution we will be standardized by performing multiple calculations to ensure the upmost accuracy. The acid used for this titration will be KHP (C4H5KO4). Phenolphthalein we be added to the beaker of the dissolved acid before the titration will be added. The titration will stop when the solution turns pink and the color doesn’t go away.
Procedure: The procedure is stated in my notebook
Data/Observations:
Observations of all trials | * The solid KHP dissolved in distilled water * The Solution turned stayed pink after enough KOH was added to the dissolved KHP. |
Trial 1: 1.5316g KHP
50.000 mL KOH in burette
28.000 mL KOH in beaker Trial 2: 1.5408g KHP
50.000mL KOH in burette
31.500mL KOH in beaker
Trial 3: 1.5380g KHP
50.000mL KOH in burette
31.500mL KOH in beaker
Analysis:
0.025L × .300M1L= .0075mol KOH
KHP (C8H5KO4) molar mass: (12.0101 ×8) + (1.00794×5) + (39.0983) + (15.9994×4) = 204.2156g
.0075mol KOH ×1mol KHP1mol KOH ×204.2156g KHP1mol KHP=1.5316g KHP for solid
1.5316g×1mole204.2156g=7.500×10-3moles 7.500×10-3moles.02800L=0.26787M
1.5408g×1mole204.2156g=7.545×10-3moles 7.545×10-3moles.03150L=0.23954M
1.538g×1mole204.2156g=7.532×10-3 7.532×10-3.03150L=0.23910M
.26787+.23954+.23910 =.74651 .746513=0.24884M KOH Trial 1: 0.26787 M Trial 2: 0.23954 M Trial 3: 0.23910 M
Accepted value: 0.24884 M Percent yield: .24884.2583×100%=96.33% Percent error:.24884-.2583.2583×100%=3.662%
Conclusion: In this experiment, a solution of KOH was standardized with titration. The KOH reacted with KHP which was