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

A, B, C

Explanation

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.