Cation groups I–V — group reagents, separation scheme: Exercises

Before starting, keep these three anchors from the parent in view:

Here $[X]$ just means "the concentration of $X$ in moles per litre" — the number of those particles crammed into one litre of solution.

Level 1 — Recognition

L1.1

State the group reagent and the precipitate type for each of Groups I, II, III, IV, V.

Recall Solution

Group	Reagent (conditions)	Precipitate type
I	dilute $H C l$	chlorides
II	$H_{2} S$ in acidic medium (dil. $H C l$ )	sulphides
III	$N H_{4} C l + N H_{4} O H$	hydroxides
IV	$H_{2} S$ in basic medium ( $N H_{4} O H$ )	sulphides
V	$(N H_{4})_{2} C O_{3}$ (ammoniacal)	carbonates

Cross-check with the table in the parent note.

L1.2

A student adds dilute $H C l$ to an unknown solution and sees a white precipitate. Which analytical group does the cation belong to, and name the three candidate cations.

Recall Solution

Group I. A white chloride precipitate points to one (or more) of $Pb^{2 +}$ , $Ag^{+}$ , $Hg_{2}^{2 +}$ . These three have chlorides with tiny $K_{s p}$ , so they fall out first with the mildest reagent. Confirmation of which one is done later — see Confirmatory Tests for Cations.

Level 2 — Application

L2.1

Explain, using anchor formula 2, why $H_{2} S$ in acidic medium precipitates Group II but not Group IV.

Recall Solution

In acidic medium $[H^{+}]$ is large. Since $[S^{2 -}] = K_{a 1} K_{a 2} [H_{2} S] / [H^{+}]^{2}$ , a large $[H^{+}]$ (squared, in the denominator) makes $[S^{2 -}]$ tiny.

With a tiny $[S^{2 -}]$ , the ionic product $Q = [M^{2 +}] [S^{2 -}]$ is small. It only manages to exceed $K_{s p}$ for sulphides whose $K_{s p}$ is extremely small — these are the Group II sulphides ( $C u S$ , $C d S$ , …). The Group IV sulphides have larger $K_{s p}$ , so their $Q$ stays below $K_{s p}$ and they remain dissolved. The whole selectivity is bought by keeping $[S^{2 -}]$ low with acid. See Weak Acid Dissociation — H2S.

L2.2

Why is $N H_{4} C l$ added together with $N H_{4} O H$ in Group III? Answer in terms of the common-ion effect.

Recall Solution

$N H_{4} O H$ is a weak base: $N H_{4} O H ⇌ N H_{4}^{+} + O H^{-}$ . Adding $N H_{4} C l$ floods the solution with $N H_{4}^{+}$ , a common ion. By Le Chatelier Principle, the extra $N H_{4}^{+}$ pushes this equilibrium left, suppressing dissociation and keeping $[O H^{-}]$ low — this is the Common Ion Effect.

A low, controlled $[O H^{-}]$ is just enough to make $Q > K_{s p}$ for the very insoluble hydroxides of $F e^{3 +}, A l^{3 +}, C r^{3 +}$ (Group III), but not enough to precipitate the higher- $K_{s p}$ hydroxides of $C o^{2 +}, N i^{2 +}, M n^{2 +}, Z n^{2 +}$ (Group IV). Selectivity, again.

Level 3 — Analysis

L3.1

A solution contains $Ag^{+}$ , $Bi^{3 +}$ , $Al^{3 +}$ , $Zn^{2 +}$ , $Ba^{2 +}$ . List, in order, the reagent added at each stage and the precipitate obtained.

Recall Solution

Follow $I \to V$ :

dil. $H C l$ → $A g C l ↓$ (white). $Ag^{+}$ is Group I.
$H_{2} S$ / acidic → $B i_{2} S_{3} ↓$ (dark brown). $Bi^{3 +}$ is Group II (small $K_{s p}$ sulphide).
$N H_{4} C l + N H_{4} O H$ → $A l (O H)_{3} ↓$ (white gelatinous). $Al^{3 +}$ is Group III.
$H_{2} S$ / basic → $Z n S ↓$ (white/dirty white). $Zn^{2 +}$ is Group IV.
$(N H_{4})_{2} C O_{3}$ → $B a C O_{3} ↓$ (white). $Ba^{2 +}$ is Group V.

One cation per group — a clean sweep.

L3.2

A student passes $H_{2} S$ before adding $H C l$ to a mix of $Ag^{+}$ and $Cu^{2 +}$ (the solution happens to be acidic). What goes wrong with the separation, and what do they see?

Recall Solution

Both a black precipitate ( $A g_{2} S$ and $C u S$ together) may form, because acidic $H_{2} S$ can precipitate the Group I sulphide $A g_{2} S$ (its $K_{s p}$ is even smaller than $A g C l$ ) alongside the Group II sulphide $C u S$ . The Group I chloride step, which should have cleanly removed silver first, is skipped — so $A g^{+}$ and $C u^{2 +}$ come out mixed and cannot be told apart.

Fix: add $H C l$ first to pull $A g^{+}$ out as white $A g C l$ , then pass $H_{2} S$ to get $C u S$ . Order matters because each group's selectivity depends on the previous group already being gone.

Level 4 — Synthesis

L4.1

Given $[C u^{2 +}] = 0.01 M$ , $K_{s p} (C u S) = 6 \times 1 0^{- 36}$ , saturated $[H_{2} S] = 0.1 M$ , $K_{a 1} = 1 \times 1 0^{- 7}$ , $K_{a 2} = 1 \times 1 0^{- 13}$ , and acidic $[H^{+}] = 0.3 M$ : compute $[S^{2 -}]$ and the ionic product $Q$ , and decide whether $C u S$ precipitates.

Recall Solution

Step 1 — sulphide level. $[S^{2 -}] = \frac{K _{a 1} K _{a 2} [ H _{2} S ]}{[ H ^{+} ] ^{2}} = \frac{( 1 0 ^{- 7} ) ( 1 0 ^{- 13} ) ( 0.1 )}{( 0.3 ) ^{2}} = \frac{1 0 ^{- 21}}{0.09} \approx 1.11 \times 1 0^{- 20} M .$ Notice how minuscule this is — that is the "tiny tail" the parent warned about.

Step 2 — ionic product. $Q = [C u^{2 +}] [S^{2 -}] = (0.01) (1.11 \times 1 0^{- 20}) = 1.11 \times 1 0^{- 22} .$

Step 3 — compare. $Q \approx 1.11 \times 1 0^{- 22}$ versus $K_{s p} = 6 \times 1 0^{- 36}$ . Since $Q ≫ K_{s p}$ (bigger by about $1 0^{14}$ ), $C u S$ precipitates — exactly as expected for Group II, even in strong acid.

L4.2

Using the same acidic conditions as L4.1, check $Ni^{2 +}$ (Group IV) with $[N i^{2 +}] = 0.01 M$ and $K_{s p} (N i S) = 3 \times 1 0^{- 21}$ . Does $N i S$ precipitate here? Then find roughly the $[H^{+}]$ at which it would just begin.

Recall Solution

Same $[S^{2 -}] = 1.11 \times 1 0^{- 20}$ M (pH unchanged). $Q = [N i^{2 +}] [S^{2 -}] = (0.01) (1.11 \times 1 0^{- 20}) = 1.11 \times 1 0^{- 22} .$ Compare with $K_{s p} (N i S) = 3 \times 1 0^{- 21}$ : now $Q < K_{s p}$ (smaller by a factor $\sim 27$ ), so $N i S$ does not precipitate in acid. Good — Group IV correctly stays dissolved.

Threshold $[H^{+}]$ . $N i S$ just begins when $Q = K_{s p}$ , i.e. $[N i^{2 +}] [S^{2 -}] = 3 \times 1 0^{- 21}$ , so $[S^{2 -}]_{needed} = \frac{3 \times 1 0 ^{- 21}}{0.01} = 3 \times 1 0^{- 19} M .$ Invert anchor 2 for $[H^{+}]$ : $[H^{+}]^{2} = \frac{K _{a 1} K _{a 2} [ H _{2} S ]}{[ S ^{2 -} ]} = \frac{1 0 ^{- 21}}{3 \times 1 0 ^{- 19}} = 3.33 \times 1 0^{- 3} \Rightarrow [H^{+}] \approx 0.0577 M .$ So dropping $[H^{+}]$ from $0.3$ M to about $0.058$ M (raising pH from $\approx 0.5$ to $\approx 1.24$ ) starts bringing $N i S$ down. In practice we go fully basic with $N H_{4} O H$ to slam $[S^{2 -}]$ up and precipitate all of Group IV — see the figure.

Figure — Cation groups I–V — group reagents, separation scheme

Level 5 — Mastery

L5.1

$P b^{2 +}$ shows up in both Group I and Group II. Using $K_{s p} (P b C l_{2}) = 1.7 \times 1 0^{- 5}$ , and a solution with $[P b^{2 +}] = 0.05 M$ before adding acid, find the residual $[P b^{2 +}]$ left in solution once $[C l^{-}] = 0.30 M$ (from the dil. $H C l$ ). Explain why this residue reappears in Group II.

Recall Solution

$P b C l_{2}$ dissolves as $P b C l_{2} ⇌ P b^{2 +} + 2 C l^{-}$ , so $K_{s p} = [P b^{2 +}] [C l^{-}]^{2}$ . Once precipitation has settled with excess $C l^{-}$ fixed at $0.30$ M, the leftover lead is $[P b^{2 +}] = \frac{K _{s p}}{[ C l ^{-} ] ^{2}} = \frac{1.7 \times 1 0 ^{- 5}}{( 0.30 ) ^{2}} = \frac{1.7 \times 1 0 ^{- 5}}{0.09} \approx 1.89 \times 1 0^{- 4} M .$ So even after Group I, about $1.9 \times 1 0^{- 4}$ M lead stays dissolved (compare with the tiny residual an ion like $A g^{+}$ leaves, since $K_{s p} (A g C l) \sim 1 0^{- 10}$ is far smaller). $P b C l_{2}$ 's $K_{s p}$ is only moderate, so its chloride is partly soluble. That surviving $P b^{2 +}$ then meets $H_{2} S$ in Group II and precipitates as black $P b S$ (whose $K_{s p}$ is vanishingly small). Hence lead is confirmed in two places — a feature, not a bug.

L5.2 (degenerate / limiting case)

What happens to $[S^{2 -}]$ as $[H^{+}] \to 0$ (a strongly basic limit)? Does the formula $[S^{2 -}] = K_{a 1} K_{a 2} [H_{2} S] / [H^{+}]^{2}$ literally hold there, and what physical cap intervenes?

Recall Solution

Mathematically, as $[H^{+}] \to 0$ the denominator $[H^{+}]^{2} \to 0$ , so the ratio blows up toward infinity. That would say $[S^{2 -}]$ is unbounded — clearly unphysical.

The escape is that the formula assumes a constant saturated $[H_{2} S] \approx 0.1$ M, which only holds while most sulphide sits as undissociated $H_{2} S$ . In strong base, $H_{2} S$ becomes almost fully deprotonated: the pool converts to $H S^{-}$ and $S^{2 -}$ , so $[H_{2} S]$ is no longer $0.1$ M — it collapses. The real cap on $[S^{2 -}]$ is the total sulphide added (conservation of mass): $[H_{2} S] + [H S^{-}] + [S^{2 -}] =$ constant. So $[S^{2 -}]$ rises sharply as we go basic (which is why Group IV precipitates), but it saturates at the total sulphide present, not infinity. This is the domain of validity behind anchor formula 2.

L5.3

A solution contains only $Mg^{2 +}$ , $Na^{+}$ , $K^{+}$ . You run the full I→V scheme. Predict the observations at every stage.

Recall Solution

These are the "Group Zero / VI" cations that the parent lists as staying in solution.

dil. $H C l$ → no precipitate ( $M g C l_{2}$ , $N a C l$ , $K C l$ all soluble).
$H_{2} S$ / acidic → no precipitate (no low- $K_{s p}$ sulphide).
$N H_{4} C l + N H_{4} O H$ → no precipitate; $M g (O H)_{2}$ stays dissolved because $N H_{4} C l$ keeps $[O H^{-}]$ low (that is the whole point — $M g (O H)_{2}$ 's $K_{s p}$ is not small enough to beat this suppressed $[O H^{-}]$ ).
$H_{2} S$ / basic → no precipitate.
$(N H_{4})_{2} C O_{3}$ → no precipitate (these carbonates are soluble / suppressed).

All three remain in solution and are found by flame tests and special reagents (e.g. magneson for $M g^{2 +}$ ), not by group precipitation. See Confirmatory Tests for Cations.

Recall One-line self-check summary

Every question above reduces to one move ::: compare the ionic product $Q$ against $K_{s p}$ under the anion concentration that the group reagent (and pH/common-ion tuning) sets.