Analysis of cations using sodium thiosulphate

Question

A question in an exam was as follows:^[1]

How many of the following salts are first soluble in $\ce{Na2S2O3}$ solution, followed by the formation of a white precipitate and finally the precipitate turns black on standing

$\ce{AgNO3}$, $\ce{Pb(CH3COO)2}$, $\ce{Hg(NO3)2}$, $\ce{BiCl3}$, $\ce{CuSO4}$

The answer given is:

$5$, which means that all of them give this test.

I understand that this is a test for cations since the reagent is sodium thiosulphate.

A pre-emptive guess on what the reactions would be that the first reaction forms the thiosulphate of the cation followed by the conversion of thiosulphates into sulphides (from the color change given).

However, I was unable to find any text referencing this test and so haven't been able to verify the actual reactions that take place.^[1]

What cations does this test give a positive result for? What are the reactions that take place?

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[1]: The previous version of the question asked for a test using sodium dithionite which was a typo (due to reasons provided in the answer below), confirmed after M.Farooq's suggestion to ask the professor who set the exam.

Answer 1

I do not think the question had sodium dithionite. This is a mistake. It must be sodium thiosulfate, because sodium dithionite is such a strong reducing agent that it will not solubilize these metals. The properties mentioned in the answer match with thiosulfate. Sodium dithionite is such a strong reducing agent that it will reduce copper (II) to copper.

With this correction... Interesting question but requires some knowledge of colors of sulfides. Sometimes you would have to use intuition in such exam questions. The hint is "...finally turn black on standing". Recall that silver tarnish is brown black, so are lead, copper and bismuth sulfides. Mercury also forms black sulfide. So in a state of exams, a safe bet would be all cations.

Thiosulfate has complex properties (a) it is a strong reducing agent (b) it forms complexes and (c) it decomposes under various conditions to pure sulfur and sulfite ion in acidic medium.

From Jander and Blasius, Textbook of Analytical and Preparative Inorganic Chemistry (in German)

Von den Salzen sind nur $\ce{BaS2O3}$, $\ce{Ag2S2O3}$ und $\ce{PbS2O3}$ schwer löslich. Sie werden jedoch durch Sodalösung quantitativ in das lösliche Natriumsalz überführt. Mit vielen Schwermetallionen (z.B. $\ce{Ag(I)}$, $\ce{Fe(II)}$ und $\ce{Cu (I)}$) bildet Thiosulfat im Überschuss lösliche Komplexe.

An english translation of this would be

Of the salts only $\ce{BaS2O3}$, $\ce{Ag2S2O3}$ and $\ce{PbS2O3}$ are poorly soluble. However, they are quantitatively converted into the soluble sodium salt by soda solution. With many heavy metal ions (e.g. $\ce{Ag(I)}$, $\ce{Fe(II)}$ and $\ce{Cu(I)}$) an excess of thiosulfate forms soluble complexes.

In the book the equations of conversion to sulfide are given as. Ask your teacher if this a typo in the exam.

4. Nachweis als $\ce{AgS2O3/Ag2S}$

\begin{align} \ce{S2O3^2- +2Ag+ &-> Ag2S2O3\downarrow} \\ \ce{Ag2S2O3 + 3S2O3^2- &-> 2[Ag(S2O3)2]^3-} \\ \ce{Ag2S2O3 + H2O &-> Ag2S + H2SO4} \\ \end{align}

That is all you have to remember that thiosulfate will precipitate $\ce{Ba}$, $\ce{Ag}$ and $\ce{Pb}$ initially but excess would dissolve it.

I would be interested to know otherwise.