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Calculate the solubility at 25^(@)C of C...

Calculate the solubility at `25^(@)C` of `CaCO_(3)` in a closed container containing a solution of `pH 8.60. [K_(sp)(CaCO_(3)) = 10^(-8)]`

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First method by using direct formula
`S_("buffer") = sqrt(K_(sp)) = [1+([H^(oplus)])/(K_(a))]^(1//2) ..(i)`
`= (10^(-8))^(1//2) [1+(2.51xx10^(-9))/(4.7xx10^(-11))]^(1//2)`
`= (54.4 xx 10^(-8))^(1//2) = 7.376 xx 10^(-4)~~ 7.4 xx 10^(-4)`.
Second method (proceeding as in the last illustration) By electroneutrality,
`[Ca^(2+)] = [CO_(3)^(2-)] + [HCO_(3)^(Θ)], ([HCO_(3)^(Θ)])/([CO_(3)^(2-)]) = 53.4`
Let `x = [Ca^(2+)] = 54.4 [CO_(3)^(2-)]`
`K_(sp) = [Ca^(2+)] [CO_(3)^(2-)]`
`= x((x)/(54.4)) = (x^(2))/(54.4) = 10^(-8)`
`:.x = 7.4 xx 10^(-4) = [Ca^(2+)]` = Solubility
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