When solid `SrCO_(3)` is equilibrated with a `pH 8.60` buffer, the solution was found to have `[Sr^(2+)] = 2.2 xx 10^(-4)`. What is the `K_(sp)` of `SrCO_(3)`.
`(K_(2)` of `H_(2)CO_(3) = 4.7 xx 10^(-11))`

Explanation

First Method by using direct formula
Solubility of salt of weak acid `(H_(2)CO_(3))` in buffer (refersection) is given by the formula
`S_("buffer") = sqrt(K_(sp)) [1+([H^(oplus)])/(K_(a))]^(1//2) ….(i)`
`S_("buffer") = 2.2 xx 10^(-4)M, pH = 8.60`,
`:.[H^(oplus)] =- log (8.6) = 2.51 xx 10^(-9)`
`K_(2) = 4.7 xx 10^(-11)`
Substituting the value in equation (i), we get
`2.2 xx 10^(-4) = (K_(sp))^(1//2) [1+(2.51xx10^(-9))/(4.7xx10^(-11))]^(1//2)`
`=(K_(sp))^(1//2) (1+53.4)^(1//2)`
`:.K_("sp") = ((2.2 xx 10^(-4))^(2))/(54.4) = 8.9 xx 10^(-10)`
Second method:
`K_(sp) = [Sr^(2+)] [CO_(3)^(2-)]`
`CO_(3)^(2-) + H_(2)O hArr HCO_(3)^(Θ) + overset(Θ)OH`
`K_(h) = (K_(w))/(K_(2)) = ([overset(Θ)OH][HCO_(3)^(Θ)])/([CO_(3)^(2-)] ) = (K_(w))/(4.7 xx 10^(-11))`
`[H_(3)O^(oplus)] =- (log 8.6) = 2.5 xx 10^(-9)`.
`[overset(Θ)OH] = (K_(w))/(2.51 xx 10^(-9))`
`((K_(w))/(2.51xx10^(-9)))[(HCO_(3)^(Θ))/(CO_(3)^(2-))] = (K_(w))/(4.7xx10^(-1))`
`([HCO_(3)^(Θ)])/([CO_(3)^(2-)]) = (2.51 xx 10^(-9))/(4.7 xx 10^(-11)) = 53.4`
The `CO_(3)^(2-)` ion which dissolves forms `HCO_(3)^(Θ)` in a `1:1` mol ratio of remains unreacted, thus, by electroneutrality,
`[Sr^(2+)] = [HCO_(3)^(Θ)] + [CO_(3)^(2-)]`
`2.2 xx 10^(-4) = 53.4 [CO_(3)^(2-)] + [CO_(3)^(2-)]`
`2.2 xx 10^(-4) = 54.4 [CO_(3)^(2-)]`
`:.[CO_(3)^(2-)] = (2.2xx10^(-4))/(54.4)`
`K_(sp) = (2.2 xx 10^(-4)) ((2.2xx10^(-4))/(54.4)) = 8.9 xx 10^(-10)`.