Investigating Stoichiometry with Carbonic Acid Salts
Investigating Stoichiometry with Carbonic Acid Salts
Problem:
Testing the actual yield versus the theoretical yield of NaCl when HCl is titrated into Na2CO3 and NaHCO3. When 0.15g of both NaHCO3 and Na2CO3 are titrated with HCl, then 0.165g of NaCl should form from the NaHCO3, and 0.104g of NaCl should form from the 0.15g of Na2CO3.
Procedure:
Weigh 2 samples of 0.15g of dried unknown each, and dissolve each into 50mL of distilled water.
Add 0.5 to 1mL of bromocresol green indicator until the solution turns blue.
Titrate the HCl until it turns green.
Gently heat and boil out the CO2. It should turn blue again.
Continue the titration until it turns yellow.
Heat the solution and let the water evaporate.
Weigh the glassware with the solid formed at the bottom, NaCl.
Dispose of the NaCl, rinse and dry the glassware. Weigh the empty glassware and record.
Find the difference from 5 and 6.
Do this for both flasks.
Results:
In table 1.1, the mixture of HCl and unknown sample 1 were combined. It took about 19.5mL of HCl to turn the solution yellow, after heating and evaporating the water from the solution, the glassware with the NaCl weighed 92.84g. When the glassware was weighed again after the NaCl was washed out, it was 92.69g. There was 0.15g of NaCl that had formed.
In table 1.2, the mixture of HCl and unkown sample 2 were combined. It took 40mL of HCl to turn the solution yellow, after the heating and evaporation of the water, the glassware with the NaCl weighed 76.38g. When the glassware was weighed again after the NaCl was washed out, it was 76.24g. There was 0.14g of NaCl that had formed.
Discussion:
The results for example 2 might have been off, when titrating the solution we added too much HCl and turned the solution yellow instead of green. So instead of boiling the solution until it turned blue again, we had to boil the solution to evaporate the water. That may have thrown off the amount of NaCl that was supposed to form after the water
Problem:
Testing the actual yield versus the theoretical yield of NaCl when HCl is titrated into Na2CO3 and NaHCO3. When 0.15g of both NaHCO3 and Na2CO3 are titrated with HCl, then 0.165g of NaCl should form from the NaHCO3, and 0.104g of NaCl should form from the 0.15g of Na2CO3.
Procedure:
Weigh 2 samples of 0.15g of dried unknown each, and dissolve each into 50mL of distilled water.
Add 0.5 to 1mL of bromocresol green indicator until the solution turns blue.
Titrate the HCl until it turns green.
Gently heat and boil out the CO2. It should turn blue again.
Continue the titration until it turns yellow.
Heat the solution and let the water evaporate.
Weigh the glassware with the solid formed at the bottom, NaCl.
Dispose of the NaCl, rinse and dry the glassware. Weigh the empty glassware and record.
Find the difference from 5 and 6.
Do this for both flasks.
Results:
In table 1.1, the mixture of HCl and unknown sample 1 were combined. It took about 19.5mL of HCl to turn the solution yellow, after heating and evaporating the water from the solution, the glassware with the NaCl weighed 92.84g. When the glassware was weighed again after the NaCl was washed out, it was 92.69g. There was 0.15g of NaCl that had formed.
In table 1.2, the mixture of HCl and unkown sample 2 were combined. It took 40mL of HCl to turn the solution yellow, after the heating and evaporation of the water, the glassware with the NaCl weighed 76.38g. When the glassware was weighed again after the NaCl was washed out, it was 76.24g. There was 0.14g of NaCl that had formed.
Discussion:
The results for example 2 might have been off, when titrating the solution we added too much HCl and turned the solution yellow instead of green. So instead of boiling the solution until it turned blue again, we had to boil the solution to evaporate the water. That may have thrown off the amount of NaCl that was supposed to form after the water