Question

Determine the molar solubility for pb3(po4)2 in pure water. ksp for pb3(po4)2 is 1.0 x 10-54.

Answer 1

Dissociation of Pb₃(PO₄)₂ is;
                       Pb₃(PO₄)₂(s)     ⇆      3Pb²⁺(aq)     +     2PO₄³⁻(aq)
initial                                                    -                          -
change               -X                            +3X                     +2X
Equilibrium                                          3X                        2X

Ksp            =       [Pb²⁺(aq)]³ [PO₄³⁻(aq)]²
1.0 x 10⁻⁵⁴ =         (3X)³ (2X)²
1.0 x 10⁻⁵⁴ =        108X⁵
         X      =        6.21 x 10⁻¹² M

Hence the molar solubility of Pb₃(PO₄)₂ is 6.21 x 10⁻¹² M.

Answer 2

The solubility product constant (Ksp) determines the solubility of the product. The molar solubility of Lead(II) phosphate is 6.21 with trillion as of the power of ten.

What is molar solubility?

Molar solubility is the moles of the solute dissolved in a liter of solution till it becomes saturated. The balanced chemical dissociation of lead phosphate is given as,

$\rm Pb_{3}(PO_{4})_{2}(s) \leftrightharpoons 3Pb^{2+}(aq) + 2PO_{4}^{3-}(aq)$

The concentration at the change of the reactant Lead(II) phosphate is -x, lead is 3x, and phosphate is 2x. At equilibrium, the concentrations of the products are the same as the change.

Ksp is calculated as:

$\begin{aligned} \rm Ksp &=\rm [Pb^{2+}]^{3} [PO_{4}^{3-}]\\\\1.0 \times 10^{-54} &= \rm (3X)^{3} (2X)^{2}\\\\&= 6.21 \times 10^{-12}\;\rm M\end{aligned}$

Therefore, the molar solubility of the solution is $6.21 \times 10^{-12} \;\rm M.$

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