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, $9.92\times 10^{-11}M$

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'

$K_{sp}=(s)\times (s)$

$K_{sp}=s^2$

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

$9.84\times 10^{-21}=(s)^2$

$s=9.92\times 10^{-11}M$

Therefore, the molar solubility of $AlPO_4$ is, $9.92\times 10^{-11}M$