Question

Given that the ksp of aipo4 is 9.84 x 10-21 at 25°c, calculate the molar solubility of aiop4

Answer 1

AlPO4----> Al+3 + PO4-3

Ksp= [Al+3] x [PO4-3]= 9.84 x 10^-21

Ksp= (x) (x)= x^2

X^2= 9.84x10-21

x= 9.92 x 10^-11

The molar solubility is 9.92 x 10^-11

Answer 2

Answer : The molar solubility of
`AlPO_4` is,
`9.92* 10^(-11)M`

Explanation :

The balanced equilibrium reaction will be,

`AlPO_4\rightleftharpoons Al^(3+)+PO_4^(3-)`

The expression for solubility constant for this reaction will be,

`K_(sp)=[Al^(3+)][PO_4^(3-)]`

Let the solubility will be, 's'

`K_(sp)=(s)* (s)`

`K_(sp)=s^2`

Now put the value of
`K_[sp}` in this expression, we get the molar solubility of
`AlPO_4`.

`9.84* 10^(-21)=(s)^2`

`s=9.92* 10^(-11)M`

Therefore, the molar solubility of
`AlPO_4` is,
`9.92* 10^(-11)M`