A coloured solution, known to contain two metal ions, was treated with excess cold sodium hydroxide solution. When filtered a whitish solid, slowly changing to brown, was retained on the filter paper and a colourless solution collected as the filtrate. Dropwise addition of hydrochloric acid to the filtrate produced a white precipitate which dissolved in excess acid. Treatment of the residue from the filter paper with a solution of a strong oxidiser produced a reddish-violet solution.
Indicate any pairs of ions which on testing as above leads to the observed changes.
- A Zn2+ and Mn2+ ions
- B Mg2+ and Zn2+ ions
- C Mn2+ and Mg2+ ions
- D Fe2+ and Zn2+ ions
- E Mn2+ and Fe2+ ions
The Attempt at a Solution
"Treatment of the residue from the filter paper with a solution of a strong oxidiser produced a reddish-violet solution." I am pretty sure this indicates Mn2+ is present as I think it turns purple. This cancels out B and D. This means that the second cation when treated with excess NaOH becomes a whitish solid, slowly changing to brown. I think this is Fe as when exposed to air it rusts and turns brown. I am pretty sure this logic is wrong though. Could someone give me some advice?