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A solution holds seven cations at once: lead(II), copper(II), iron(III), zinc, nickel(II), barium…
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A solution holds seven cations at once: lead(II), copper(II), iron(III), zinc, nickel(II), barium and magnesium. An analyst adds the group reagents in order: dilute hydrochloric acid, hydrogen sulphide in dilute acid, ammonium chloride and ammonium hydroxide, hydrogen sulphide in ammonium hydroxide, ammonium carbonate, and last the test for magnesium. Each time the precipitate is removed. Lead is removed completely in the first step. Find four numbers. First, the cations precipitated by hydrogen sulphide in dilute acid. Second, the cations left in solution after the group III reagent. Third, the cations precipitated by ammonium carbonate. Fourth, the cations that give a coloured solution in water. The coloured ones are copper(II) (blue), iron(III) (yellow-brown) and nickel(II) (green).
Need a reminder? The key idea
The worked solution
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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.
Step 1. Group I removes lead: 6 left.
Step 2. Hydrogen sulphide in dilute acid: copper only, so 1 precipitated and 5 left.
Step 3. Group III: iron(III) only, so 4 are left (zinc, nickel, barium, magnesium).
Step 4. Group IV: zinc and nickel are removed, leaving barium and magnesium. Group V: barium is removed by ammonium carbonate, so 1.
Step 5. Coloured: copper(II), iron(III), nickel(II): 3.
Answer: 1, 4, 1 and 3.
Check. The removed numbers are , so every cation was found.
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