Question

The solubility of $Ca_3(PO_4)_2$ in water is y moles / litre. Its solubility product is

• A. $6y^4$
• B. $36y^4$
• C. $64y^5$
• D. $108y^5$

Answer 1

Answer: (D)

$Ca_3(PO_4)_2(s)\rightleftharpoons \underset{\text{3y}}{3Ca^{2+}} (aq) +\underset{\text{2y}}{2PO_4^{3-}}(aq) $
$\therefore K_{sp}=\left[Ca^{2+}\right]^{3}\cdot\left[PO_{4}^{3-}\right]^{2}$
$=\left(3y\right)^{3}\cdot\left(2y\right)^{2}$
$=27y^{3}\times4y^{2}$
$=108\,y^{5}$