A student dissolves Mohr's salt, FeSO₄· (NH₄)₂SO₄· ₆H₂O, in water and adds sodium hydroxide…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A student dissolves Mohr's salt, , in water and adds sodium hydroxide. A green solid forms. She repeats it with potassium hexacyanidoferrate(III), , and no solid forms. Say which compound is a double salt, and explain why the two solutions behave differently. How many ions does one formula unit of each compound give in water?
Need a reminder? The key ideaShow
Werner's theory: the primary valence is the oxidation state, and the secondary valence is the coordination number. The coordination sphere is the metal with its ligands, in square brackets. Only the counter ions outside the brackets are free in water, so only they give a precipitate with silver nitrate. A double salt gives all its ions. A complex keeps its complex ion.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Plan. A double salt falls into the ions of its parts. A complex keeps its complex ion.
Step 1. Mohr's salt. It is a double salt. In water it gives one ion, two ions and two ions, which is 5 ions. The free ions give the green solid, which is iron(II) hydroxide.
Step 2. The cyanido compound. It is a coordination compound. In water it gives three ions and one ion, which is 4 ions. The iron stays inside the complex ion, so there are no free iron ions and no solid.
Answer: Mohr's salt is the double salt and gives 5 ions. gives 4 ions.
Check. Both have iron, but only Mohr's salt has iron that is free in water.
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