4. For a solution containing at least one of the cations involved in this experi

Question

4. For a solution containing at least one of the cations involved in this experiment, answer the following questions. (1) A solution is acidified to pH0.5, and HS is added. A dark precipitate results. Which cation(s) may be present? (2) KOH is added to the precipitate from (1). The supernatant liquid is separated from the dark precipitate and tested with HCI/H2S. A colorless solution results, with no precipitate. Which cation(s) may be present? (3) The remaining precipitate from (1) is treated with dilute HCl. The supernatant liquid is tested with NH' and H2S. A yellow precipitate results. Which cation(s) may be present? (4) The precipitate left from the first step in (3) is dissolved in HNOs, and the resulting supernatant liquid is treated with NH3. A blue supernatant liquid results, with no precipitate. Give the identity of all Group II cations present in the original solution.

Explanation / Answer

(1) If dark precipitate obtained on addition of H2S in an acidic solution of cations, there might have following cations

Pb2 +, Cu2 +, Bi3 +, Hg2+, Cd2+, As3+, Sb3+, and Sn4+ as the S2- salt of this cations appeared as dark-colored and forms in acidic solution.

(2) Upon addition of KOH only can dissolve Sb and Sn sulphide as they form following ions and soluble in water.

SbS33-, SbO33-, SnS33-, SnS2OH-

Then on the addition of HCl corresponding Chlorides will form as follows,

SbS33- + HCl = SbCl4- + H2S, SbO33- + HCl = SbCl4- + H2O,

SnS33- + HCl = SnCl62- + H2S, SnS2OH-+ HCl = SnCl62- + H2O + H2S

So they are solutble in water. Now when you add H2S to this solution if it is remains colourless then it would be Sn4+ because if it would Sb3+ then it must form orange colored SbS3.

(3) Again the precipitate from (1) upon treatment with HCl will dissolve the sulfide salt by forming corresponding chloride salts of the respective metal. This may contain Pb2+ or Cd2+.

Since H2S additions give a yellow precipitate that may be because of formation of CdS(yellow). So the present metal ion is Cd2+.

(4) HNO3 further dissolves CuS and Bi2S3 into the water. The blue coloration of supernatant liquid on the addition of ammonia is because of the presence of Cu2+because it forms Cu(NH3)42+ complex which has the blue color.

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