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, $3.32\times 10^{-209}$

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'.

$k_{sp}=(3s)^3\times (2s)^2$

$k_{sp}=108\times s^5$

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

$k_{sp}=108\times (7.9\times 10^{-43})^5=3.32\times 10^{-209}$

Therefore, the value of $K_{sp}$ of $Pb_3(PO_4)_2$ is, $3.32\times 10^{-209}$