The solubility of Ca_3(PO_4)_2 in water is y | vedclass

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Question

(D) The dissociation of $Ca_3(PO_4)_2$ is given by:
$Ca_3(PO_4)_2{_{	ext{(s)}}} ightleftharpoons 3Ca^{2+}{_{	ext{(aq)}}} + 2PO_4^{3-}{_{	ext{(aq

Vedclass · Accepted Answer

$108 y^5$

Vedclass · Answer

(D) The dissociation of $Ca_3(PO_4)_2$ is given by:
$Ca_3(PO_4)_2{_{	ext{(s)}}} ightleftharpoons 3Ca^{2+}{_{	ext{(aq)}}} + 2PO_4^{3-}{_{	ext{(aq)}}}$
If the solubility is $y 	ext{ mol/L}$, then the concentration of ions at equilibrium is:
$[Ca^{2+}] = 3y$
$[PO_4^{3-}] = 2y$
The solubility product expression is:
$K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2$
Substituting the values:
$K_{sp} = (3y)^3 (2y)^2$
$K_{sp} = (27y^3) (4y^2)$
$K_{sp} = 108y^5$

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

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