The yellow solution remaining after Group A separation and containing possible G

Question

The yellow solution remaining after Group A separation and containing possible Group B cations is treated as described in the lab procedure. At each stage below indicate what the following tests tell you about which cations may be present, which are confirmed present, which are definitely absent, and why.

(a) Without any treatment, the solution has a yellow color.

(b) The solution from step (a) is treated with HC1 and NaHCO3, centrifuged, and separated to give the following to give the following results.

1. A white precipitate is obtained.

2. An orange solution is obtained.

(b) The precipitate from step (b) is dissolved in HC1 and treated with aluminum metal and then Hg2+ to give a white solid.

(d) The orange solution from step (b) is divided into two portions and each treated separately as follows.

1. One portion is treated with HC1, aluminon, and NH3 giving a clear red solution with no precipitate.

2. The second portion is treated with 3% H2O2 and HC1. The solution turns dark blue and quickly fades to colorless.

Explanation / Answer

(a) The unknown solution is made acidic with HCl, the following reactions that may occur are as follows :-

Al(OH)-1(aq) + 4H+1(aq) Al+3(aq) + 4H2O(l)

Sn(OH)-16(aq) + 6H+1(aq) + 6Cl-1(aq) SnCl-2(aq) + 6H2O(l)

2CrO-24(aq) (yellow) + 2H+1(aq) Cr2O7-2(aq) (orange) + H2O(l)

The yellow colour of solution may be due to the presence of CrO-24 .

(b) The solution is then treated with NaHCO3 to neutralize the acid. Then centrifuged.

(1) The white precipitate is obtained. :- Tin (IV) ion precipitates as SnO2.

SnCl-26(aq) + 2H2O(l) + 4NH3(aq) SnO2(s)(white precipitate) + 4NH+14(aq) + 6Cl-1(aq)

Therefore, Sn+4 ion is present.

(2) An orange solution is obtained. :- The solution may contain chromium (III) ion, which can be conclude from the following reaction -

2CrO-24(aq) (yellow) + 2H+1(aq) Cr2O7-2(aq) (orange) + H2O(l)

(c)  The precipitate from step (b) is dissolved in HCl and treated with aluminum metal and then Hg+2 to give a white solid. :-

White precipitate of SnO2 is treated with HCl to dissolve the precipitate. Aluminum metal is then added to convert the solid tin(II) oxide to SnCl4-2.

SnO2(s) + 4H+1(aq) + 6Cl-1(aq) SnCl6-2(aq) + 2H2O(l)

3SnCl6-2(aq) + 2Al(s) 3SnCl4-2(aq) + 2Al+3(aq) + 6Cl-1(aq)

When aluminum metal is added to an acidic solution, bubbling occur due to the formation of hydrogen gas.

2Al(s) + 6H+l(aq) 2Al+3(aq) + 3H2(g)

Mercury (II) is then added which oxidizes Sn+2 back to Sn+4. Hg+2 is reduced to Hg2+2, mercury(I), which precipitates as the white-colored Hg2Cl2, Sn+2 is confirmed.

SnCl4-2(aq) + 2Hg+2(aq) + 4Cl-1(aq) Hg2Cl2(s) (white) + SnCl6-2(aq)

(d) The orange solution from step (b) is divided into two portions :-

.1. One portion is treated with HCl, aluminon, and NH3 giving a clear red solution with no precipitate. :-

As the solution when treated with red aluminon, doesn't gives precipitate, Al+3 is absent.

2. The second portion is treated with 3% H2O2 and HCl. The solution turns dark blue and quickly fades to colorless. :-

The test for chromium involves reduction of dichromate ion by hydrogen peroxide in acidic solution to give the blue CrO5 species. CrO5 is unstable and the blue color fades rapidly. Therefore, Cr+3 is present.

Cr2O7-2(aq) + 4H2O2(aq) + 2H+1(aq) 2CrO5(aq) (blue) + 5H2O(l)

CrO5(aq) + 6H+l(aq) Cr+3(aq) + O2(aq) + 3H2O(l)

The fleeting appearance of a blue color confirms Cr+3.

From the above test, it is concluded that the given solution contains the cations of metal Sn and Cr.

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