Question

a) A Group III unkown is colorless. What cations are probably absent? Why should one not rely definitely upon such an observation?

b) Why will PbCrO₄ precipiate when Pb²⁺ is added to a solution made up from K₂Cr₂O₇?

Answer 1

a)

Group III cations, which include Cr3+, Al3+, Fe3+ and Ni2+

Fe3+, Ni2+ and Cr3+ are absent because these are forms colored Precipitates.

An aqueous mixture of Cr3+, Al3+, Fe3+ and Ni2+ is first treated with a mixture of NaOH and NaOCl solutions. This causes the iron and nickel cations to precipitate out as hydroxide salts, while the chromium and aluminum cations remain in solution: Fe3+ (aq) + 3 OH- (aq) → Fe(OH)3 (s) Ni2+ (aq) + 2 OH- (aq) → Ni(OH)2 (s)

2 Cr3+ (aq) + 3 OCl- (aq) + 10 OH- (aq) → 2 CrO4 2- (aq) + 3 Cl- (aq) + 5 H2O (l)

Al3+ (aq) + 4 OH- (aq) → Al(OH)4 - (aq)

Note in the last two equations above that Cr3+ is oxidized to the soluble CrO4 2- ion while Al3+ forms a soluble complex ion with OH- . Neither iron nor nickel form hydroxo-complex ions and therefore precipitate out as solids. The resulting mixture is centrifuged and then decanted, separating the solids (Fe(OH)3 and Ni(OH)2) from the supernatant solution (containing CrO4 2- and Al(OH)4 - ). In order to separate the chromium ions from the aluminum ions in the aqueous supernatant solution, the solution is first acidified in order to destroy the aluminum hydroxo-complex ion:

Al(OH)4 - (aq) + 4 H+ (aq) → Al3+ (aq) + 4 H2O (l)

b)

Most lead compounds are insoluble (except for acetates, perchlorates, chlorates, and nitrates). Therefore, when the soluble salts lead(II) nitrate and potassium chromate are mixed, insoluble lead(II) chromate forms and precipitates out (K_sp = 2.8 x 10^-13).

¹Pb(NO₃)₂(aq) + K₂CrO₄(aq) ==> PbCrO₄(s) + 2 KNO₃(aq) or Pb²⁺(aq) + CrO₄^2-(aq) ==> PbCrO₄(s) Lead(II) chromate is also known as "chrome yellow" .