A Cobalt-Amine-Halide compound is synthesized from cobalt (II) chloride hexahydrate. An orange-tinted solid is produced and is considered to be unknown since the specific ligand amounts are unknown. By determining the percent composition of various elements and compounds in the unknown, its true identity can be predicted. Chloride, ammonia, and cobalt are three examples of percent compositions determined to help narrow the selection of possible unknowns. Titrations using Na2S2O3 and HCl to determine percent cobalt and ammonia, respectively, are used.
Silver nitrate is used to precipitate the chloride ions in the unknown, which can be measured to determine the percent composition of chloride in the unknown. The results from these three major analyses helped to draw the conclusion that the unknown is in fact [Co(NH3)6]Cl3. -2. 0 Synthesis/Analysis of Cobalt-Amine-Halide Compund (Procedure 1, pgs 16-7)) A cobalt (III) amine halide compound was prepared by making a solution of cobalt (II) chloride hexahydrate and ammonium chloride (NH4Cl) to liberate the Co (II) in the solution. “Activated” charcoal and ammonia are then added to the solution.
The Co (II) is then oxidized to Co (III) using hydrogen peroxide (H2O2).
This yields an orange crystalline solid of 8. 9379 grams, later found to be [Co(NH3)6]Cl3. [Co(H2O)6]+2 + 6 NH3 ? [Co(NH3)6]+2 + 6 H2O 2 [Co(NH3)6]+2 + 2 H+ + H2O2 ? 2 [Co(NH3)6]+3 + 2 H2O -3. 0 Analysis for Percent Halide In The Synthesized Cobalt-Ammonia-Halide Compound (pgs 23-4) HNO3 is used to acidify the soluble chloride-containing cobalt compound. Silver nitrate is used in excess to precipitate chloride ions (Cl-) as AgCl in an approximately . 2400 grams of the unknown compound.
The Essay on Chemistry Halide Ions
Purpose: There are four halide salts used in this experiment that are found in the human body. Sodium fluoride is poisonous, but has been traced to be beneficial to humans in the prevention of tooth decay. Sodium chloride is added to many of our foods to increase flavor. Sodium chloride is important for many life processes, but too much intake is linked to high blood pressure. Sodium bromide is ...
This is done in duplicate and purple-tinted precipitates are placed in Gooch crucibles. The precipitates are suction dried using ethyl alcohol then acetone to rinse. After being dried in the oven for thirty minutes, they are weighed to be . 3655 grams and . 3824 grams. The silver halide precipitates can be used to calculate the percent halide in the precipitate and in the unknown. The average percent halide was found to be 38. 00% with 3. 08% deviation between trials 1 and 2 (Table 1&2).
X Cl = 35. 45 g/mole Cl= . 0904g Cl- . 0904g Cl- * 100 = 36. 78% Cl . 3655g 143. 34 g/mole AgCl . 2458g Co sample Mass of Co-sample
Mass of precipitate % halide in Co-sample .2458g .3655g 36. 78% .2412g .3824g 39. 22% Average ? 38. 00% -4. 0 Preparation & Standardization Of An HCl Solution (pg 25) The amount of 12M HCl needed to prepare 500mL of 0. 3M HCl was found to be 12. 5mL. After the HCl is prepared, it is titrated in triplicate with about . 9605 grams of THAM each time. Brom-cresol-green indicator is used to determine when the solution has reached its endpoint or equilibrium. The molarity of HCl came to an average of . 3045M with a percent deviation of . 0657% (Table 3).
(HOCH2)3CNH2 + HCl ? (HOCH2)3CNH3+ + Cl- 1) (12M HCl)(V) = (0. 3M)(500mL)
V= 12. 5mL 2) 26. 00mL * 1L = . 02600L HCl 1000mL .9610g THAM * 1 mol THAM * 1 mole HCl = . 0079255 mole HCl 121. 14 g/mole THAM 1 mole THAM .0079255 mole HCl = . 3048M HCl .02600L Grams of THAM Final volume – initial volume Molarity of HCl .9610g 26. 00ml .3051M .9602g 26. 05ml .3043M .9602g 26. 05ml .3043M Averages ?. 9605g 26. 03ml .3045M -5. 0 Analysis Of Percent NH3 Using Standardized . 3M HCl (pgs 26-8) Excess of the strong base NaOH was added to about . 37 grams of the unknown cobalt compound. NH3 reacts with boric acid and is liberated. This solution was then titrated with the . 3045M HCl using about 26. 05mL.
The Essay on Solution Preparation And Standardization
Generally, there are two ways in preparing a solution, one is by dissolving a weighed amount of solid in a required solvent and the other is by dilution of a concentrated solution into the desired concentration. In diluting concentrated solution, the concentration of the diluted solution can be determined by standardization. To standardize a solution, we will need to perform titration. In this ...
An average percent NH3 of 38. 25% was found with a percent deviation of 2. 240% (Table 4).
2 Co(NH3)6Cl3 + 6 NaOH ? 2 Co(OH)3 + 12 NH3 + 6 NaCl heat + 2 Co(OH)3 ? Co2O3 + 3 H2O 12 NH3 + 12 H3BO3 ? 12 NH4+ + 12 H2BO3- 12 HCl + 12 H2BO3 ? 12 H3BO3 + 12 Cl- overall, 2 Co(NH3)6Cl3 + 6 NaOH + 12 HCl ? Co2O3 + 12 NH4+ + 12 Cl- + 6 NaCl 1) 27. 25 mL * 1 L = . 02725 L 1000 mL .02725 L * . 3046 mole HCl * 1 mole NH3 * 17. 03 g NH3 = . 1414 g NH3 1 L HCl 1 mole HCl 1 mole NH3 . 1414 g NH3 * 100 = 37. 43% NH3 .3778 g Co-compound Grams of Co(III) compound Final volume – initial volume % NH3 .3739g 27. 30ml 39. 53% .3778g 27. 25ml 37. 43%
.3735g 27. 20ml 37. 78% Averages ? .3751g 27. 25ml 38. 25% -6. 0 Preparation and Standardization of 0. 1M Sodium Thiosulfate (Na2S2O3) (pgs 29-30) A 500mL 0. 1M sodium thiosulfate solution (Na2S2O3) was created and mixed with Na2CO3, which acts as a preservative. KIO3 is added to an HCl and KI solution containing excess KI and I- ions. KIO3 acts as a primary standard and reacts to produce I2, which can then be reacted with the Na2S2O3. The yellow/brown color during the triplicate titrations is due to the presence of I3-. Starch indicator is added to the slightly titrated yellow-colored solution to help easily see the endpoint.
The liberated I2 needed about 31. 49mL of Na2S2O3 solution during the titration to reach the endpoint. The molarity of Na2S2O3 used for the titration is found to be . 10270M with . 159% deviation (Table 5).
IO3- + 6 H+ + 5 I- ? 3 I2 +3 H2O 6 S2O32- + 3 I2 ? 3 S4O62- + 6 I- overall, IO3- + 6 H+ + 6 S2O32- ? 3 S4O62 + I- + 3 H2O or KIO3 + 6 HCl + 6 Na2S2O3 ? 3 Na2S4O6 + KI + 3 H2O + 6 NaCl 1) 31. 13 mL * 1L = . 03113L 1000mL .1139 g KIO3 * 1 mole KIO3 = . 0005322 mole KIO3 214 g/mole KIO3 .0005322 mole KIO3 * 6 mole Na2S2O3 = . 0031932 mole Na2S2O3 1 mole KIO3 .0031932 mole Na2S2O3 = . 10258M Na2S2O3
.03113 L Na2S2O3 Grams of KIO3 (last row is average) Final volume – initial volume Molarity of thiosulfate solution (Na2S2O3) .1139g 31. 13ml .10258M .1147g 31. 35ml .10258M .1175g 32. 00ml .10295M .1154g 31. 49ml .10270M -7. 0 Analysis of Percent Co+3 Using Standardized 0. 1M Thiosulfate (Na2S2O3) (pgs 30-2) In triplicate, about . 5590 grams of the unknown cobalt compound is mixed with 50% NaOH and heated for twenty minutes without boiling to avoid loss of cobalt. KI is added in excess to the cooled solution then combined with 6M HCl in small increments in order to acidify the now red-brown solution.
The Essay on Osmosis Of Potatoes In Different Sucrose Solutions
The aim of this experiment is to test whether more water moves out of a potato when it is placed in a sweeter sucrose solution than a potato in a less sweet solution. The hypothesis of the experiment is that we expect more water to move out of the potato placed in the sweet solution than the potato placed in a less sweet solution. Independent variable: concentration of sucrose, concentrations: ...
The titration is partially done then starch indicator is added and changes color due to the presence of I3-. About 20. 02mL of the . 1091M Na2S2O3 solution is used to reach completion. The unknown compound therefore contains about 23. 1% Co with a 5. 19% deviation (Table 6).
2 Co(NH3)4Cl3 + 6 OH- ? 2 Co(OH)3 + 6Cl- + 8 NH3 6 H+ + 2Co(OH)3 ? 2 Co+3 + 6HOH 2 Co+3 + 2 I- ? 2 Co++ + I2 2 S2O32- + I2 ? S4O62- + 2I- overall, 2 Co(NH3)4Cl3 + 2 S2O32- ? 2 Co+2 + S4O62- + 8 NH3 + 6Cl- 1) 20. 15 mL Na2S2O3 * 1 L = . 02015 L Na2S2O3 1000 mL .02015 L Na2S2O3 * . 1091 mole Na2S2O3 * 1 mole Co+3 = . 0021984 mole Co+3 1 L Na2S2O3 1 mole Na2S2O3
.0021984 mole Co+3 * 58. 93g Co+3 = . 1296 g Co+3 1 mole Co+3 .1296 g Co+3 * 100 = 21. 51% .6025 g Co-sample Grams of Co (III) compound in grams Final volume – initial volume % Co in sample .6025 g 20. 15 mL 21. 51% .5176 g 20. 05 mL 24. 90% .5568 g 19. 85 mL 22. 92% Averages ? .5590 g 20. 02 mL 23. 11% -8. 0 Analysis for H2O2 In Commercial 3% Solutions (pgs 35-6) 1. 5 grams of KI is mixed with 20 drops of 12M HCl and swirled. Fresh H2O2 is pipetted into the solution. The flasks sit for 5 minutes then are titrated in triplicate until a yellow color is reached. Starch indicator is added and the titrated is continued to
