Solution - Career Point

Precipitation of the hydroxides of group III :Hydroxides of group III are precipitated by adding anexcess of solid NH 4 Cl to the solutions of thesecations followed by the addition of excess ofNH 4 OH. Being a weak electrolyte, NH 4 OH is onlyslightly ionised, whereas NH 4 Cl, being a strongelectrolyte, ionizes almost completely to give at largeconcentration of NH + 4 ions.NH 4 OH NH + 4 + OH – ; NH 4 Cl → NH + 4 + Cl –Due to the common ion effect, the degree ofdissociation of NH 4 OH gets suppressed and hence theconcentration of OH – ions in solution decreasesappreciably. But even with this low conc. of OH –ions, the ionic products of the cations of group III andOH – ions exceed the low values of the solubilityproducts of their corresponding metal hydroxides. Asa result, the cations of group III get precipitated astheir insoluble hydroxides.On the other hand, cations of groups IV, V and Mg,which require a large conc. of OH – ions due to theirhigh solubility products will not be precipitated.Precipitation of sulphides of group IV. Thesulphides of group IV are precipitated by passing H 2 Sthrough ammoniacal solution of these cations.Both H 2 S and NH 4 OH, being weak electrolytes,ionize only slightly as :H 2 S 2H + + S 2–+NH 4 OH NH 4 + OH –The H + ions and OH – ions combine to producepractically unionised molecules of waterH + + OH – → H 2 OAs a result, the above dissociation equilibriumreactions get shifted in the forward direction, so thatthe concentration of S 2– ions goes on increasing.Ultimately, the ionic product of the cations of groupIV and S 2– ions exceed the solubility products of theircorresponding metal sulphides and hence getprecipitated.Precipitation of carbonates of group V : Thecarbonates of group V are precipitated by adding(NH 4 ) 2 CO 3 solution to the solution of these cations inthe presence of NH 4 Cl and NH 4 OH. (NH 4 ) 2 CO 3 ,being a weak electrolyte ionises only slightly to givea small concentration of CO 2– 3 ions.(NH 4 ) 2 CO 3 2NH + 2–4 + CO 3On the other hand, NH 4 Cl being a strong electrolyte,ionises almost completely to give a large+concentration of NH 4 ions. Due to the common ioneffect, the dissociation of (NH 4 ) 2 CO 3 is suppressedand hence the concentration of CO 2–3 ions in thesolution decreases considerably. But even with thislow concentration of CO 2– 3 ions, the ionic products ofthese cations and CO 2– 3 ions exceed the low values ofthe solubility products of their corresponding metalcarbonates and thus get precipitated.However, under these conditions, Mg salts do not getprecipitated as MgCO 3 since its solubility product iscomparatively high and thus requires a highconcentration of CO 2–3 ions for precipitation. Thecarbonates of Na + , K + +and NH 4 ions are also notprecipitated because they are quite soluble.The necessity of adding NH 4 OH arises due to the factthat (NH 4 ) 2 CO 3 solution usually contains a largeamount of NH 4 HCO 3 . Thus, the cations of group Vwill form not only insoluble carbonates but solublebicarbonates as well. As a result, the precipitationwill not be complete. In order to convert NH 4 HCO 3to (NH 4 ) 2 CO 3 , NH 4 OH is always added.NH 4 HCO 3 + NH 4 OH → (NH 4 ) 2 CO 3 + H 2 OPreferential precipitation of Salts :A solution contains more than one ion capable offorming a precipitate with another ion which is addedto the solution. For example, in a solution containingCl – , Br – , and I – ions, if Ag + ions are added, then outof the three, the least soluble silver salt is precipitatedfirst. If the addition of Ag + ions is continued,eventually a stage is reached when the next lessersoluble salt starts precipitating along with the leastsoluble salt and so on. If the stoichiometry of theprecipitated salts is the same, then the salt with theminimum solubility product (and hence also theminimum solubility) will precipitate first followed bythe salt of next higher solubility product and so on.If the stoichiometry of the precipitated salts is not thesame, then, from the solubility product data alone, wecannot predict which ion will precipitate first. Take,for example, a solution containing Cl – and CrO 2– 4 .Both these ions form precipitates with Ag + . Throughthe solubility product product of AgCl is larger thanthat of Ag 2 CrO 4 , yet it is AgCl (lesser soluble) whichprecipitates first when Ag + ions are added to thesolution. In order to predict which ion (Cl – or CrO 2– 4 )precipitates first, we have to calculate theconcentration of Ag + ions needed to start theprecipitation through the solubility product data andthe given concentration of Cl – or CrO 2– 4 . Since squareroot is involved in the expression for computing Ag +for silver chromate, the quantity of Ag + needed tostart the precipitation of CrO 2– 4 is larger than that forCl – . Hence, as AgNO 3 is added to the solution, theminimum of the two concentrations of Ag + to startthe precipitation will be reached first and thus thecorresponding ion (Cl – in this case) will beprecipitated in preference to the other. During thecourse of precipitating, concentration of Cl –decreases and the corresponding concentration of Ag +to start the precipitation increases. Its concentrationeventually becomes equal to the value required forCrO 2– 4 . At this stage, practically the whole of Cl –ions have been precipitated. The addition of more ofAgNO 3 causes the precipitation of both the ionstogether.XtraEdge for IIT-JEE 33 APRIL 2010