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Giventhat the solubility product of BaSO_(4) is 1xx10^(-10). Will a precipitate form when (i) Equal volumes of 2xx10^(-3)M BaCl_(2) solution and 2xx10^(-4) M Na_(2)SO_(4) solution are mixed ? (ii) Equal volumes of 2xx10^(-8) MBaCl_(2) solution and 2xx10^(-3)M Na_(2)SO_(4) solution are mixed ? |
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Answer» Solution : (i) `BaCl_(2)` ionizes completely in the solution as : `BaCl_(2) rarr Ba^(2) + 2 Cl^(-) :. [Ba^(2+)]=[BaCl_(2)]=2xx10^(-3)M` (Given) `Na_(2)SO_(4)` ionizes completely in the solution as : `Na_(2)SO_(4) rarr 2Na^(+)+SO_(4)^(2-) :. [SO^(4)^(2-) :. [SO_(4)^(2-)] = [ Na_(2)SO_(4)]=2xx10^(-4)M` (Given) Since equal volumes of the two solutions are mixed together, therefore, the concentrations of `Ba^(2+)` ions and `SO_(4)^(2-)` ions after mixing will be `[Ba^(2+)]=(2xx10^(-3))/(2) = 10^(-3) M and [SO_(4)^(2-)]=(2xx10^(-4))/(2) = 10^(-4)M` `:. ` Ionic PRODUCT of `BaSO_(4) = [ Ba^(2+)][SO_(4)^(2-)]=10^(-3)xx10^(-4)=10^(-7)` which is greater than the solubility product `(1xx10^(-10))` of `BaSO_(4)` . Hence, a precipitate of `BaSO_(4)` will be FORMED. (ii) Here, the concentrations before mixing are : `[Ba^(2+)]=[BaCl_(2)]=2xx10^(-8) M ` (Given), `[SO_(4)^(2-)]=[Na_(2)SO_(4)]=2xx10^(-3)M ` (Given) `:.` Concentrations after mixing equal volumes willbe : `[Ba^(2+)]= (2xx10^(-8))/(2) = 10^(-8)M and [SO_(4)^(2-)] = (2xx10^(-3))/(2) = 10^(-3) M` `:.` Ionic product of `BaSO_(4) = [ Ba^(2+)][SO_(4)^(2-)]=10^(-8) xx 10^(-3)=10^(-11)` which is less than the solubility product `(1xx10^(-10))`. Hence, no ppt. will be formed in this case. |
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