Solubility of calcium phosphate (molecular ma | vedclass

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(B) The solubility $S$ in $mol \ L^{-1}$ is calculated as: $S = \frac{W 	imes 1000}{M 	imes 100} = \frac{10W}{M} \ mol \ L^{-1}$.The dissociation of

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$10^7 \left(\frac{W}{M}\right)^5$

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(B) The solubility $S$ in $mol \ L^{-1}$ is calculated as: $S = \frac{W 	imes 1000}{M 	imes 100} = \frac{10W}{M} \ mol \ L^{-1}$.The dissociation of calcium phosphate is: $Ca_3(PO_4)_{2(s)} ightleftharpoons 3Ca^{2+}(aq) + 2PO_4^{3-}(aq)$.The solubility product expression is: $K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2 = (3S)^3 (2S)^2$.$K_{sp} = 27S^3 	imes 4S^2 = 108S^5$.Substituting $S = \frac{10W}{M}$: $K_{sp} = 108 	imes \left(\frac{10W}{M}ight)^5 = 108 	imes 10^5 	imes \left(\frac{W}{M}ight)^5 = 1.08 	imes 10^7 \left(\frac{W}{M}ight)^5$.Rounding to the nearest order of magnitude,the value is approximately $10^7 \left(\frac{W}{M}ight)^5$.

Solubility of calcium phosphate (molecular mass,$M$) in water is $W \ g$ per $100 \ mL$ at $25^{\circ} C$. Its solubility product at $25^{\circ} C$ will be approximately.

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