InterviewSolution
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InterviewSolution
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The solubility of `BaSO_(4)` in water is `2.33 g 100 mL^(-1)`. Calculate the percentage loss in weight when `0.2g` of `BaSo_(4)` is washed with a. `1L` of water b. `1L` of `0.01 NH_(2)SO_(4).[Mw_(BaSO_(4)) = 233 g mol^(-1)]` |
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Answer» a. Solubility is in general expressed in `gL^(-1)`, so soubility of `BaSo_(4) = 2.33 xx 10^(-3) gL^(-1)` Loss in weight of `BaSO_(4) =` Amount of `BaSO_(4)` soluble. `rArr %` loss `= (2.33 xx 10^(-3))/(0.2) xx 100 = 1.16%` b. `0.01 NH_(2)SO_(4) -= 0.01 N SO_(4)^(2-) ions` `-= 0.0055 M SO_(4)^(2-) ion (n` factor for `SO_(4)^(2-) = 2)` Now presence of `SO_(4)^(2-)` prior to washing `BaSO_(4)` will supress the solubility of `BaSO_(4)` (due to common ion effect. ) The supersion will be governed by `K_(sp)` value of `BaSO_(4)`. So, first calculate the `K_(sp)` of `BaSO_(4)`. Solubility of `BaSO_(4)` in fresh water `= 2.33 xx 10^(-3) gL^(-1)` `= (2.33 xx 10^(-3))/(33) ,mol L^(-1) = 10^(-5)M` `K_(sp)= [Ba^(2+)][SO_(4)^(2-)] = (10^(-5))^(2) = 10^(10)` Now let `x` be the solubility in `mol L^(-1) in H_(2)SO_(4)` `rArr [Ba^(2+)] `in solution `= x mol L^(-1)` and `[SO_(4)^(2-)]` in solution `= (x + 0.005) mol L^(-1)` Ionic product `= [Ba^(2+)] [SO_(4)^(2-)] = (x) (x + 0.005)` `K_(sp) =` Ionic product at equilibrium (saturation) `rArr 1.0 xx 10^(-10) = (x) (x + 0.005)` Assuming `x` to be a small number `(x + 0.005) ~~ 0.005` `rArr x = (10^(-10))/(0.005) = 2xx10^(-8)mol L^(-1)` `= 2 xx 10^(-8) xx 233 gL^(-1)` ` = 4.66 xx 10^(-6) gL^(-1)` `rArr 4.66 xx 10^(-6)g of BaSO_(4)` is washed away. `rArr % loss = (4.66 xx 10^(-6)xx100)/(0.2) = 2.33 xx 10^(-3) %` |
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