Question

The solubility of lead iodide in water is 0.63g/litre.calculate the solubility product of lead iodide(Atomic mass – Pb =207,1=127)​

Answer 1

Answer:

Let the solubility of PbI  

2

​  

 be S. Then

PbI  

2

​  

⇌Pb  

2+

+2I  

−

 

S             S           2S

Potassium iodide is a strong electrolyte and is completely ionised. It shall provide I  

−

 ion concentration =0.1M

[Pb  

2+

]=S

[I  

−

]=(2S+0.1)M

K  

sp

​  

=[Pb  

2+

][I  

−

]  

2

=S×(2S+0.1)  

2

=4S  

3

+0.01S+0.4S  

2

 

Neglecting S  

3

 and S  

2

 

1.4×10  

−8

=0.01S

or S=  

0.01

1.4×10  

−8

 

​  

=1.4×10  

−6

mol L  

−1

Explanation: