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.
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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).
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