Briefly explain what the consequense would be if excess 6M NaOH solution instead of ^m NH3 solution were used to separate Fe3+ and Al3+ ions from Ca2+, Cu2+,and K+ ions at pH 9-10.
The idea behind NH3 is that it complexes the Cu (forms [Cu(NH3)]4^2+ and provides enough OH- to ppt Al^3+ as Al(OH)3 and Fe^3+ as Fe(OH)3 and leaves the others mentioned in solution. Using NaOH ppts the Fe as Fe(OH)3 BUT it will dissolve the Al(OH)3 to Al(OH)4^-, Cu will ppt as (Cu(OH)2, and it PROBABLY will ppt Ca^2+ as Ca(OH)2 (but I'm not positive about that since there are no concns given).
If an excess 6M NaOH solution is used instead of a ^m NH3 solution to separate Fe3+ and Al3+ ions from Ca2+, Cu2+, and K+ ions at pH 9-10, the following consequences may occur:
1. Precipitation of Fe(OH)3 and Al(OH)3: The addition of excess NaOH to the solution would result in the precipitation of Fe(OH)3 and Al(OH)3. These hydroxide precipitates are insoluble in water and would form as solids, separating them from the Ca2+, Cu2+, and K+ ions.
2. Insufficient separation of Ca2+, Cu2+, and K+ ions: Since NaOH is not a selective precipitant for Ca2+, Cu2+, and K+ ions, these ions are not efficiently separated from the Fe3+ and Al3+ ions. Consequently, some amount of Ca2+, Cu2+, and K+ ions will be present in the resulting precipitate alongside Fe(OH)3 and Al(OH)3.
3. Formation of Na+ ions: Excess NaOH would also lead to an excess of Na+ ions in the solution. These Na+ ions would remain in the solution and potentially interfere with any further analysis or separation processes.
Overall, using excess 6M NaOH solution instead of an NH3 solution would result in incomplete separation of Fe3+ and Al3+ ions from Ca2+, Cu2+, and K+ ions, leading to some contamination in the precipitate and the presence of excess Na+ ions in the solution.
If excess 6M NaOH solution instead of an NH3 solution is used to separate Fe3+ and Al3+ ions from Ca2+, Cu2+, and K+ ions at pH 9-10, it would have some consequences. Let's break it down step by step:
1. Purpose of separation: The goal of this separation process is to selectively precipitate Fe3+ and Al3+ ions as hydroxides while leaving Ca2+, Cu2+, and K+ ions in the solution. This is achieved by adjusting the pH to a specific range and using a specific precipitating agent.
2. NaOH vs. NH3: In the usual separation process, NH3 (ammonia) solution is used as the precipitating agent. NH3 forms complexes with Fe3+ and Al3+ ions, making them soluble at higher pH levels. On the other hand, NaOH (sodium hydroxide) solution causes the precipitation of hydroxides of Fe3+ and Al3+ ions.
3. Consequences of using NaOH instead of NH3:
a. Incomplete separation: NaOH is a strong base and tends to precipitate most metal hydroxides, including Ca(OH)2, Cu(OH)2, and K(OH). Therefore, using excess NaOH would result in the precipitation of Ca2+, Cu2+, and K+ ions along with Fe3+ and Al3+ ions. The separation of these ions would be incomplete, defeating the purpose of the separation process.
b. pH disturbance: The addition of excess NaOH to the solution would significantly increase the pH level. This higher pH may affect the stability of certain compounds and complexes, leading to unpredictable results and interfering with subsequent analysis or procedures.
c. Formation of unwanted precipitates: The presence of excess NaOH would cause the precipitation of hydroxides of Ca2+, Cu2+, and K+, forming unwanted precipitates that can further contaminate the separation. It would complicate the subsequent steps required to isolate Fe3+ and Al3+ ions.
In summary, using excess 6M NaOH instead of an NH3 solution in the given separation process would result in incomplete separation, disturbance of pH, and the formation of unwanted precipitates. It is important to follow the specified procedure to achieve the desired separation efficiently.