InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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))` |
|
Answer» 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)`. |
|