Question

Calculate the solubility product of (1)CuCl and (2) Ag2Cro5 the solubilities being 8.7*10*moles/litre and 7.8*10 moles/litre respectively.​

Answer 1

Answer:

Ag  

2

​

CrO  

4

​

⇌  

2S

2Ag  

+

 

​

+  

S

CrO  

4

2−

​

 

​

 

K  

sp(Ag  

2

​

CrO  

4

​

)

​

=[Ag  

+

]  

2

[CrO  

4

2−

​

]

=[2S]  

2

[S]

=4S  

3

 

=4×(8.0×10  

−5

)  

3

 

K  

sp

​

=2.048×10  

−12

 

K  

sp

​

 for K  

2

​

CrO  

4

​

 at 25  

∘

C will be 2.048×10  

−12

.