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 = 5.03×1013. what is K?

Answer 1

Answer : The value of 'K' is,
`1.78* 10^(4)`

Explanation :

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

(1) The chemical reaction will be:

`Cu_3(PO_4)_2\rightleftharpoons 3Cu^(2+)+2PO_4^(3-)`

`K_(sp)=1.40* 10^(-37)`

The copper(II) ion also forms a complex ion with ammonia.

(2) The chemical reaction will be:

`Cu^(2+)+4NH_3\rightarrow [Cu(NH_3)_4]^(2+)`

`K_f=5.03* 10^(13)`

To calculate the value of 'K' we are multiplying reaction 2 by 3, we get:

(3)
`3Cu^(2+)+12NH_3\rightarrow 3[Cu(NH_3)_4]^(2+)`

`K_f'=(K_f)^3=(5.03* 10^(13))^3`

Now we are adding reaction 1 and 3, we get:

`Cu_3(PO_4)_2+12NH_3\rightarrow [Cu(NH_3)_4]^(2+)+2PO_4^(3-)`

`K=K_f'* K_(sp)`

`K=(5.03* 10^(13))^3* (1.40* 10^(-37))`

`K=1.78* 10^(4)`

Therefore, the value of 'K' is,
`1.78* 10^(4)`