Explain why magnesium,aluminium and iron metals cannot be identified by flame test.
Answers
Answer:Metal hydroxide precipitate tests
Dilute sodium hydroxide solution is used in tests for some metal ions, which form metal hydroxides that are insoluble. This means that the metal hydroxides appear as precipitates. For example, copper sulfate solution reacts with a few drops of sodium hydroxide solution:
copper sulfate + sodium hydroxide → sodium sulfate + copper hydroxide
CuSO4(aq) + 2NaOH(aq) → Na2SO4(aq) + Cu(OH)2(s)
Copper hydroxide forms a blue precipitate.
Sodium hydroxide solution is added to copper sulfate solution. Solid copper hydroxide is produced in colourless sodium sulfate solution.
Sodium hydroxide solution is added to copper sulfate solution. Solid copper hydroxide is produced in sodium sulfate solution
The table shows the coloured precipitates formed by five common metal ions.
Aluminium, Al3+White
Calcium, Ca2+White
Magnesium, Mg2+White
Copper(II), Cu2+Blue
Iron(II), Fe2+Green
Iron(III), Fe3+Brown
Distinguishing between aluminium ions, calcium ions and magnesium ions
A few drops of dilute sodium hydroxide solution react to form a white precipitate with aluminium ions, calcium ions and magnesium ions. However, if excess sodium hydroxide solution is added:
the aluminium hydroxide precipitate dissolves to form a colourless solution
the calcium hydroxide precipitate is unchanged
the magnesium hydroxide solution is unchanged
This means that using sodium hydroxide can give a positive result for aluminium ions, but it cannot distinguish between calcium and magnesium ions.
Question
A green precipitate forms when dilute sodium hydroxide solution is added to a sample in solution. Identify the metal ion present in the original solution.
Reveal answer
The ionic equations
The precipitation reactions can be represented using ionic equations, which only include the ions which are involved in the formation of the precipitate. They ignore the spectator ions, which are present but not involved. For example:
Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s)
In this equation, we can see that:
the copper ion has a charge of 2+
therefore two hydroxide ions are needed to balance this because they each have a charge of -1
therefore the formula of the hydroxide precipitate has two OH ions in it
In the precipitation reaction which identifies the iron(III) ion as being present, the ionic equation is:
Fe3+(aq) + 3OH-(aq) → Fe(OH)3(s)
This is because three hydroxide ions are needed to react with the Fe3+ ion.
The spectator ions which are ignored in these equations are the sodium ion (Na+) from the NaOH, and the anion from the metal compound, eg the sulfate ion (SO42-) if the copper compound was copper sulfate.(︶^︶)(-_-)→_→(=_=)=_=