The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferential precipitation. It is the solubility of the salt,not the solubility product of the salt, by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example , in a solution containing `Cl^(-), Br^(-)` and `I^(-)` ions, if `Ag^(+)` ions are added, then out of three , the less soluble salt is precipitated first . If the addition of `Ag^(+)` ions is continued, eventually , a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on.
If the stiochiometry of the salt is same, then the salt with the minimum `K_(sp)` ( solubility product ) will have minimum solubility and will precipitate first, followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same, then from the solubility product data alone, we cannot predict which ions will precipitate first. For example, a solution containing `Cl^(-)` and `CrO_(4)^(2-)`. In order to predict which ion will precipitate first, we have to calculate the amount of `Ag^(+)` ions needed to start precipitation through the solubility product data given. When `AgNO_93)` is added to the solution, the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding `[Ag^(+)]` to start the precipitation increases. Its concentration eventually becomes equal to the value required for `CrO_(4)^(2-)` . At this stage, practically almost the whole of `Cl^(-)` ions get precipitated . Addition of more `AgNO_(3)` causes simultaneous precipitation of both the ions together.
Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount.
Experiment 1. He prepare a solution of anions composed of 0.1 M each of `Cl^(-),Br^(-)` and `I^(-)` . Further he adds gradually solid `AgNO_(3)` to this solution. He assumes that volume of the solution does not change after the addition of solid `AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2),K_(sp) AgBr=10^(-10)M^(2)` and `K_(sp) AgI=10^(-12)M^(2))`
Experiment -2. He prepares a solution of cations `Cd^(2+)(0.2 M)` and `Bi^(3+) (0.3 m)` . Now he adds `S^(2-)` ions into the solution of cations in order to separate them by selective precipitation `Cd^(2+) `forms yellow precipitate of CdS and `Bi^(3+)` forms black precipitate of `Bi_(2)S_(3)` with `S^(2-)` ions respectively.
`(K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5)` and `K_(sp)CdS =2 xx 10^(-20)M^(2))`.Answer the following questions ont he basis of the above write up.
What percent of the anions `Br^(-)` and `I^(-)` get precipitated respectively when the third ion starts precipitating in experiment -1 ?
A. `90%,99.9%`
B. `99.9%,90%`
C. `80%,90%`
D. `90%,80%`
