An unknown solution containing Group A and Group B cations is treated according

Question

An unknown solution containing Group A and Group B cations is treated according to the lab procedure. At each stage below, state what each test tells you about which cations may be present, which are confirmed present, which are absent, and why.

Group A: Bi3+, Fe3+, Mn2+   Group B: Al3+, Cr3+, Sn4+

(a) The solution is treated with NH3 until the pH is 9 to give a colored precipitate. The precipitate was treated with NaOH-H2O2 and the mixture centrifuged and separated to give a yellow solution and a precipitate. The yellow solution was saved for the Group B cation analysis. The remaining precipitate was dissolved in HCl and the clear solution analyzed as follows.

(b) Four drops of the clear solution from step (a) are treated with NaBiO3 to give a purple solution.

(c) Two drops of the solution from step (a) are treated with 6M NaOH and solid SnC12 to give a black solid.

(d) Aportionofthesolutionfromstep(a)istreatedwithKSCNandthesolutionremainsclear.

Explanation / Answer

Groups A and B cations react with NH3 (at pH 10) and precipitate as hydroxides or oxides.

Group A cations react NH3 (at pH 10)   according to the following equations;

Bi+3(aq) + 3 NH3(aq) + 3 H2O(l) Bi(OH)3(s) (white) + 3 NH4+1(aq)

Fe+3(aq) + 3 NH3(aq) + 3 H2O(l) Fe(OH)3(s) (red-brown) + NH4+1(aq)

Mn+2(aq) + 2 NH3(aq) + 2 H2O(l) Mn(OH)2(s) (pale pink) + 2 NH4+1(aq)

Sumilarly , Group B cations precipitate as Al(OH)3 (white ), Cr(OH)3 (green) and SnO2 (white).

On addition of NaOH and H2O2, to the mixed group precipitate, the Group B cations will dissolve while the Group A cations remain as solids. At this point, the Group A cations can be separated physically from Group B by centrifuging and decanting. Hence the seperation of group A from group B.

ppt of Bi(OH)3 and Fe(OH)3 do not react further with either NaOH or H2O2, but Mn(OH)2 is oxidized to MnO2 as shown in the equation below.

Mn(OH)2(s) + H2O2(aq) MnO2(s) + 2H2O(l)

The precipitate of Group A cations is dissolved in hot HCl to give a solution containing Bi+3, Fe+3, and Mn+4 ions. Some Mn+4 may be converted to Mn+2 but this will have no effect on the confirmation of manganese.

b. Manganese(II) Ion Test: when Mn+4 ion treated with H2O2 it convert to Mn+2. The Mn(II) ion react with bismuthate ion (BiO3 -1) to form the purple permanganate ion. The appearance of the purple permanganate color confirms Mn+2.

Mn+4(aq) + H2O2(aq) Mn+2(aq) + O2(aq) + 2H+l(aq)

14H+l(aq) + 2Mn+2(aq)+ 5 BiO3-1(aq) 5Bi+3(aq)+ 7H2O(l) + 2MnO4-1(aq) (purple)

c. Bismuth Ion test: Bi+3 ion, present as Bi(OH)3, is reduced to metallic bismuth (Bi0 ) by Sn+2. This reaction occurs in basic solution in which Sn(II) exists as the Sn(OH)3-1 ion.

The appearance of a black precipitate confirms Bi+3.

2 Bi+3(aq) + 30H-l(aq) Bi(OH)3(s)

Sn+2(aq) + OH-1(aq) Sn(OH)3-1(aq)

2Bi(OH)3(s) + 3Sn(OH)3-1(aq) + 3 OH-1(aq) 2 Bi(s) + 3 Sn(OH)6-2(aq) (black)

d. iron test :

Fe+3(aq) + SCN-l(aq) FeSCN+2(aq) (red-brown)

we get a clear solution, no red brown coluor present, indicates absence of Fe3+

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.