The result of mixing solutions containing Cu(H2O)6 2+ and NH4+ can be thought of

Question

The result of mixing solutions containing Cu(H2O)6 2+ and NH4+ can be thought of as Cu 2+ and H+ competing for the lone pair on NH3. That is, both Cu 2+ and H+ react to a large extent. If a reaction were to occur between Cu(H2O)6 2+ and NH4+ ions, where Cu 2+ won the competition with H+ for NH3, what would the equation be? The result of mixing solutions containing Cu(H2O)6 2+ and NH4+ can be thought of as Cu 2+ and H+ competing for the lone pair on NH3. That is, both Cu 2+ and H+ react to a large extent. If a reaction were to occur between Cu(H2O)6 2+ and NH4+ ions, where Cu 2+ won the competition with H+ for NH3, what would the equation be? If a reaction were to occur between Cu(H2O)6 2+ and NH4+ ions, where Cu 2+ won the competition with H+ for NH3, what would the equation be?

Explanation / Answer

The equation would be :-

[Cu(H2O)6]2+ + NH4+ -----> [Cu(NH3)2(H2O)4]2+ + 2H2O

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