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Given that solubility product of `BaSO_(4)` is `1 xx 10^(-10)` will be precipiate from when a. Equal volumes of `2 xx 10^(-3)M BaC1_(2)` solution and `2 xx 10^(-4)M Na_(2)SO_(4)` solution, are mixed? b. Equal volumes of `2 xx 10^(-8) M BaC1_(2)` solution and `2 xx 10^(-3)M Na_(2)SO_(4)` solution, are mixed? c. `100mL` of `10^(-3)M BaC1_(2)`and `400mL` of `10^(-6)M Na_(2)SO_(4)` are mixed. |
Answer» Correct Answer - A::B::C i. `BaC1_(2)` ionise completely in the solution as `BaC1_(2) rarr Ba^(2+) + 2C1^(Theta)` `[Ba^(2+)] = [BaC1_(2)] = 2 xx 10^(-3)M` (given) `Na_(2)SO_(4)` ionise completely in the solution as `Na_(2)SO_(4) rarr 2Na^(o+) + SO_(4)^(2-)` ` :. [SO_(4)^(2-)] = [Na_(2)SO_(4)] = 2 xx 10^(-4)M` (given) Since equal volume of the two solution are mixed together, the concentration 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) xx 10^(-4) = 10^(-7)` which is greater than the solubility product `(1 xx 10^(-10))` of `BaSO_(4)`. Hence, a precipitate of `BaSO_(4)` will be formed. ii. Here, the concentration before mixing is: `[Ba^(2+)] = [BaC1_(2)] = 2 xx 10^(-8)M` `[SO_(4)^(2-)] = [Na_(2)SO_(4)] = 2 xx 10^(-3)M` (Given) Concentration after mixing equal volumes will be `[Ba^(2+)] = (2 xx 10^(-8))/(2) = 10^(-8)M` `[SO_(4)^(2-)] = (2 xx 10^(-3))/(2) = 10^(-3)M` `:.` Ionic product of `BaSO_(4) = [Ba^(2+)] [SO_(4)^(2-)]` `= 10^(-8) xx 10^(-3) = 10^(-11)` whcih is less than the solubility product `(1 xx 10^(-10))`. Hence, no ppt will be formed in this case. iii. Concentration of `Ba^(2+)` (Total volume `= 100 + 400 = 500M`) `{:(M_(1)V_(1),=,M_(2)V_(2),,),(("Before"),,("After"),,):}` `100 xx 10^(-3) = M_(2 xx 500` `M_(2) = (100 xx 10^(-3))/(500) = (10^(-3))/(5)M` Concentration of `SO_(4)^(2-)` `{:(M_(1)V_(1),=,M_(2)V_(2),,),(("Before"),,("After"),,):}` `400 xx 10^(-6) = M_(2) xx 500` `M_(2) = (400 xx 10^(-6))/(500) = (4)/(5) xx 10^(-6)M` `IP (or) Q_(sp)` of `BasO_(4) = [Ba^(2+)] [SO_(4)^(2-)]` ` = (10^(-3))/(5) xx (4)/(5) xx 10^(-6)` `(4)/(25) xx 10^(-9) M^(-2)` `= 0.16 xx 10^(-9)M^(2) = 1.6 xx 10^(-10)M^(2)` `Q_(sp) gt K_(sp)` So precipitate of `BaSO_(4)` will taken there. |
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