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  4. In a saturated solution of calcium phosphate, the concentration of PO43- ions is 3.3 × 10-7 M. The Ksp of Ca3(PO4)2 will be

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Q. In a saturated solution of calcium phosphate, the concentration of $PO_4^{3-}$ ions is $3.3 × 10^{-7}\, M$. The $K_{sp}$ of $Ca_3(PO_4)_2$ will be

Equilibrium

Solution:

$Ca_3(PO_4)_4 \rightarrow 3Ca^{2+} +2PO^{3-}4$
$\left[Ca^{2+}\right]=\frac{3}{2}\left[PO^{3-}_{4}\right]=\frac{3}{2}\times\left(3.3\times10^{-7}\right)M$
$=4.95\times10^{-7}\,M$
$K_{sp}=\left[Ca^{2+}\right]^{3}\left[PO^{3-}_{4}\right]^{2}$
$=\left(4.95\times10^{-7}\right)^{3}\times\left(3.3\times10^{-7}\right)M$
$=4.95\times10^{-7}\,M$
$K_{sp}=\left[Ca^{2+}\right]^{3}\left[PO^{3-}_{4}\right]^{2}$
$=\left(4.95\times10^{-7}\right)^{3}\times\left(3.3\times10^{-7}\right)^{2}=1.32\times10^{-32}$