The copper(II) ion reacts with phosphate ion to form an insoluble ionic compound

Question

The copper(II) ion reacts with phosphate ion to form an insoluble ionic compound, Ksp = 1.40×10^-37. The copper(II) ion also forms a complex ion with ammonia, [Cu(NH3)4]2+, Kf = 5.03×10^13.

Write the chemical reaction that occurs between aqueous ammonia and solid copper(II) phosphate.

Hint: The ammonia will react with the small amount of free copper(II) ions present and form the complex ion [Cu(NH3)4]2+. We can start by writing the reactions corresponding to Ksp and Kf. By combining these reactions we can derive the reaction that will occur between copper(II) ion and aqueous ammonia.

Explanation / Answer

Solution:- Cu3(PO4)2(s) <-----> 3Cu2+(aq) + 2PO43-(aq)

Ksp = [Cu2+]3 [PO43-]2

3Cu2+(aq) + 12NH3(aq) <----> 3[Cu(NH3)4]2+(aq) (need to multiply the second equation 3 to cancel out the copper ions.)

Kf = ([Cu(NH3)4]2+)3/[Cu2+]3[NH3]12

Over all equation :- Cu3(PO4)2(s) +  12NH3(aq) <-----> 3[Cu(NH3)4]2+(aq) + 2PO43-(aq)

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