Question

The molar solubility of lead(ii) phosphate (pb3(po4)2) is 7.9 x 10-43 m. calculate the solubility product constant (ksp) of lead(ii) phosphate.

Answer 1

the answer is 3.3 x 10^-209 . i hope this helps.

Answer 2

Answer : The value of
`K_(sp)` of
`Pb_3(PO_4)_2` is,
`3.32* 10^(-209)`

Explanation :

The equilibrium reaction will be,

`Pb_3(PO_4)_2\rightleftharpoons 3Pb^(2+)+2PO_4^(2-)`

The expression of solubility product,
`K_(sp)` will be,

`k_(sp)=[Pb^(2+)]^3[PO_4^(2-)]^2`

Let the molar solubility be 's'.

`k_(sp)=(3s)^3* (2s)^2`

`k_(sp)=108* s^5`

Now put the value of 's' in this expression, we get :

`k_(sp)=108* (7.9* 10^(-43))^5=3.32* 10^(-209)`

Therefore, the value of
`K_(sp)` of
`Pb_3(PO_4)_2` is,
`3.32* 10^(-209)`