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

Before anything, let us pin down two words we will lean on constantly.

Building the sulphide formula we keep reusing

Several cells lean on one equation, so we earn it here rather than dropping it from the sky.

Formula Deriving

$[S^{2 -}]$ — shown, not asserted We want $[S^{2 -}]$ in terms of things we control ( $[H^{+}]$ and the roughly-constant $[H_{2} S]$ ). Why? Because precipitation depends on $[S^{2 -}]$ , but we can only tune the acidity.

Step 1 — multiply the two equilibria. Multiplying the two $K$ expressions lets the awkward $H S^{-}$ cancel: $K_{a 1} K_{a 2} = \frac{[ H ^{+} ] [ H S ^{-} ]}{[ H _{2} S ]} \cdot \frac{[ H ^{+} ] [ S ^{2 -} ]}{[ H S ^{-} ]} = \frac{[ H ^{+} ] ^{2} [ S ^{2 -} ]}{[ H _{2} S ]} .$ Why multiply? The $[H S^{-}]$ appears once on top, once on bottom — it divides out, leaving only species we care about.

Step 2 — solve for $[S^{2 -}]$ . We now isolate $[S^{2 -}]$ by moving $[H^{+}]^{2}$ across and multiplying by $[H_{2} S]$ : $[S^{2 -}] = \frac{K _{a 1} K _{a 2} [ H _{2} S ]}{[ H ^{+} ] ^{2}}$ Why solve for $[S^{2 -}]$ specifically, and in this form? Because $[S^{2 -}]$ is the one quantity that decides precipitation ( $Q = [M^{2 +}] [S^{2 -}]$ ), yet it is invisible and un-adjustable directly. This form expresses it purely through two knobs we can set — the acidity $[H^{+}]$ and the (fixed) saturation level $[H_{2} S]$ — so every later cell just substitutes numbers here. The $[H^{+}]^{2}$ sits in the denominator, so $[S^{2 -}] \propto 1/ [H^{+}]^{2}$ : more acid ⇒ far less sulphide. Figure 1 (below Cell A) draws the separation flowchart that this control ultimately powers.

Now the numeric power of this shows up in Cell G, and its direction in Cell B.

The scenario matrix

Every worked example is tagged with the cell of this matrix it covers, so you can see the coverage is complete.

Cell	Case class	The awkward thing it tests
A	Clean multi-cation trace	ordinary I→V routing, no traps
B	Sign / direction of pH	does raising or lowering $[H^{+}]$ precipitate?
C	Borderline / "in two groups"	$P b^{2 +}$ leaks from I into II
D	Degenerate: only one cation	scheme must still not mis-fire
E	Zero / absent reagent	skip HCl — what breaks?
F	Limiting value: exact threshold	$Q = K_{s p}$ , the precipitate-or-not knife-edge
G	Numeric $[S^{2 -}]$ from pH	plug real numbers into the master formula
H	Real-world word problem	tap-water hardness (Group V logic)
I	Exam twist	common-ion suppression changes the answer

Cell A — a clean multi-cation trace

Worked example Cell A. Solution has

$A g^{+}$ , $B i^{3 +}$ , $C r^{3 +}$ , $M n^{2 +}$ , $S r^{2 +}$ . Route each to its group. Forecast: one ion per group — guess which group each lands in before reading.

Add dilute HCl. $A g C l ↓$ (white). Why this step? HCl is the most selective reagent (only three chlorides have a tiny enough $K_{s p}$ ). $A g^{+}$ is one of them → Group I.
Pass $H_{2} S$ in acidic medium. $B i_{2} S_{3} ↓$ (dark brown). Why? Acidic medium keeps $[S^{2 -}]$ tiny (see Cell G), so only very-low- $K_{s p}$ sulphides drop. $B i^{3 +}$ → Group II.
Add $N H_{4} C l + N H_{4} O H$ . $C r (O H)_{3} ↓$ (green). Why? $N H_{4} C l$ holds $[O H^{-}]$ low (common-ion), just enough for the tiniest- $K_{s p}$ hydroxides. $C r^{3 +}$ → Group III.
Pass $H_{2} S$ in basic medium. $M n S ↓$ (buff/pink). Why? Now $[S^{2 -}]$ is large, so higher- $K_{s p}$ sulphides finally exceed $K_{s p}$ . $M n^{2 +}$ → Group IV.
Add $(N H_{4})_{2} C O_{3}$ . $S r C O_{3} ↓$ (white). Why? Only carbonates of the alkaline-earths are left insoluble. $S r^{2 +}$ → Group V.

Verify: the order I→II→III→IV→V is strictly reagent-selectivity ascending. No ion appears twice, so no cross-contamination — a valid clean trace. Confirm each with Confirmatory Tests for Cations.

The whole routing logic — one reagent per stage, in order — is drawn as a flowchart (Figure 1):

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

Cell B — which direction of pH precipitates?

Worked example Cell B. You have

$N i^{2 +}$ dissolved with $H_{2} S$ bubbling through, currently acidic. To precipitate $N i S$ , do you add acid or add base? Forecast: more $H^{+}$ or less $H^{+}$ ? Guess.

Write the sulphide supply: $[S^{2 -}] = \frac{K _{a 1} K _{a 2} [ H _{2} S ]}{[ H ^{+} ] ^{2}}$ (derived above). Why this step? $N i S$ needs $Q = [N i^{2 +}] [S^{2 -}] > K_{s p}$ ; the only knob is $[S^{2 -}]$ , and this formula tells us how it responds to $[H^{+}]$ .
Note the $[H^{+}]^{2}$ in the denominator. Why it matters: $[S^{2 -}] \propto 1/ [H^{+}]^{2}$ . So lowering $[H^{+}]$ (adding base) raises $[S^{2 -}]$ .
$N i^{2 +}$ is Group IV (larger $K_{s p}$ ) — it needs a big $[S^{2 -}]$ . So we must add base ( $N H_{4} O H$ ). Adding acid would drive $[S^{2 -}]$ down and keep $N i S$ dissolved.

Verify: direction check — halving $[H^{+}]$ multiplies $[S^{2 -}]$ by $4$ (since $1/ (0.5)^{2} = 4$ ). The dependence is inverse-square, so the sign is unambiguous: base precipitates Group IV. This is Le Chatelier Principle — removing $H^{+}$ pulls both dissociation steps forward.

Cell C — the borderline ion that shows up twice

Worked example Cell C. A solution contains lead at total concentration

$[P b^{2 +}]_{total} = 0.05$ M (a typical analytical level) and nothing else. After dilute HCl you still find lead in the Group II sulphide test. Contradiction? Forecast: is this an error, or expected? Guess.

$P b C l_{2}$ has a moderate $K_{s p} \approx 1.7 \times 1 0^{- 5}$ — not tiny like $A g C l$ . Why this step? Group-I membership needs a very small $K_{s p}$ ; $P b C l_{2}$ only just qualifies, so some lead will survive in solution.
Fix the $[C l^{-}]$ from the reagent. "Dilute HCl" here is about $0.1$ M HCl added in modest excess, so the free chloride settles near $[C l^{-}] \approx 0.1$ M. Why this assumption? The chloride is supplied overwhelmingly by the added HCl, not by the small amount of lead that stays dissolved; so we may treat $[C l^{-}] \approx 0.1$ M as fixed by the reagent, independent of how much $P b C l_{2}$ dissolves. Our $0.05$ M starting lead is far more than enough to saturate $P b C l_{2}$ at this chloride level (as Step 3 confirms), so solid $P b C l_{2}$ is present and the solution is genuinely saturated.
Now the dissolved lead left in solution (in equilibrium with solid $P b C l_{2}$ ) is $[P b^{2 +}] = \frac{K _{s p}}{[ C l ^{-} ] ^{2}} = \frac{1.7 \times 1 0 ^{- 5}}{( 0.1 ) ^{2}} = 1.7 \times 1 0^{- 3} M .$ Why compute this? It quantifies the residue — the lead that Group I fails to remove. Since $1.7 \times 1 0^{- 3}$ M is far below the $0.05$ M we started with, most lead did precipitate as $P b C l_{2}$ , but a real, measurable tail stays dissolved.
That residual $P b^{2 +}$ then meets $H_{2} S$ in Group II; $P b S$ has a minuscule $K_{s p} \approx 1 0^{- 28}$ , so it precipitates easily. → lead detected again.

Verify: the residual $1.7 \times 1 0^{- 3}$ M is over a thousand times higher than the near-zero left by a truly Group-I ion like $A g^{+}$ , yet a small fraction ( $1.7 \times 1 0^{- 3} /0.05 = 3.4%$ ) of the starting lead. So finding $P b^{2 +}$ in both Group I and Group II is expected, not a mistake. (Use a larger excess of HCl, $[C l^{-}] = 0.3$ M, and the residue falls to $1.7 \times 1 0^{- 5} /0.09 \approx 1.89 \times 1 0^{- 4}$ M — still non-trivial. Borderline $K_{s p}$ = split personality.)

Cell D — the degenerate single-cation case

Cell E — the zero-reagent case (skip HCl)

Worked example Cell E. Solution has

$A g^{+}$ and $C u^{2 +}$ . A careless student skips HCl and goes straight to acidic $H_{2} S$ . What goes wrong? Forecast: which ions precipitate, and can they be told apart?

Acidic $H_{2} S$ gives a low but non-zero $[S^{2 -}]$ . Why this step? Both $A g_{2} S$ ( $K_{s p} \approx 6 \times 1 0^{- 51}$ ) and $C u S$ ( $K_{s p} \approx 8 \times 1 0^{- 37}$ ) have insanely small $K_{s p}$ , so even a tiny $[S^{2 -}]$ makes $Q > K_{s p}$ for both.
So a mixed black precipitate of $A g_{2} S + C u S$ forms together. Why it's bad: Group I ( $A g^{+}$ ) was supposed to be removed first as $A g C l$ ; now silver and copper are entangled.
You have lost the I–vs–II separation entirely.

Verify: Group I exists precisely to pull $A g^{+}$ , $P b^{2 +}$ , $H g_{2}^{2 +}$ out before any sulphide step. Setting "reagent = zero" (no HCl) breaks the ordering guarantee. This is exactly the "add $H_{2} S$ first" mistake in the parent note — the matrix's zero-input cell.

Cell F — the exact threshold ( $Q = K_{s p}$ )

Worked example Cell F.

$[Z n^{2 +}] = 0.01$ M, and $[S^{2 -}]$ is tuned so that $Q$ exactly equals $K_{s p} (Z n S) = 1.6 \times 1 0^{- 24}$ . What $[S^{2 -}]$ is that, and does $Z n S$ precipitate? Forecast: at $Q = K_{s p}$ — solid or no solid?

Set $Q = K_{s p}$ : $[Z n^{2 +}] [S^{2 -}] = 1.6 \times 1 0^{- 24}$ . Why this step? The knife-edge is defined by equality; solving it locates the exact boundary (the amber dot in Figure 2).
Solve for the unknown $[S^{2 -}]$ by dividing both sides by the known $[Z n^{2 +}]$ : $[S^{2 -}] = \frac{1.6 \times 1 0 ^{- 24}}{0.01} = 1.6 \times 1 0^{- 22}$ M. Why divide by $[Z n^{2 +}]$ ? Because $[Z n^{2 +}]$ is the concentration we fixed and $[S^{2 -}]$ is what we want; isolating it tells us the exact sulphide level that sits on the boundary — anything more precipitates, anything less dissolves.
At $Q = K_{s p}$ the solution is exactly saturated — no net precipitation. Solid forms only when we push $[S^{2 -}]$ above $1.6 \times 1 0^{- 22}$ M (i.e. $Q > K_{s p}$ ).

Verify: $0.01 \times 1.6 \times 1 0^{- 22} = 1.6 \times 1 0^{- 24} = K_{s p}$ ✓. The limiting case is the boundary: below it dissolves, above it precipitates, at it nothing changes. Every group cut-off is really "which side of this equality is each ion on."

The single comparison $Q$ vs $K_{s p}$ that drives every cell is drawn as a number line (Figure 2), with beakers P and Q from Cell I placed on either side of the amber $K_{s p}$ mark:

Cell G — real numbers: $[S^{2 -}]$ from pH, and the II/IV cut

Worked example Cell G. Saturated

$[H_{2} S] = 0.1$ M, $K_{a 1} K_{a 2} = 1.1 \times 1 0^{- 21}$ . Compute $[S^{2 -}]$ at (i) $[H^{+}] = 0.3$ M (acidic, Group II) and (ii) $[H^{+}] = 1 \times 1 0^{- 9}$ M (ammoniacal, Group IV). Which sulphides precipitate at $[M^{2 +}] = 0.01$ M? Forecast: guess the ratio of the two $[S^{2 -}]$ values (hint: it's a power).

Where does $[H_{2} S] = 0.1$ M come from? $H_{2} S$ gas is only weakly soluble; when we bubble it through until the solution can hold no more, the water saturates at about $0.1$ mol L $^{- 1}$ at room temperature and $1$ atm. Because $H_{2} S$ is barely ionised (tiny $K_{a 1}$ ), almost all of that stays as undissociated $H_{2} S$ , so its concentration holds near $0.1$ M throughout the test — see Weak Acid Dissociation — H2S. That is why we treat $[H_{2} S]$ as a fixed constant in the master formula.
Master formula: $[S^{2 -}] = \frac{K _{a 1} K _{a 2} [ H _{2} S ]}{[ H ^{+} ] ^{2}}$ . Why? This is the sulphide level we derived above — the whole II/IV split lives here. (Note $K_{a 1} K_{a 2} = 1 0^{- 7} \times 1.1 \times 1 0^{- 14} = 1.1 \times 1 0^{- 21}$ , matching the given product.)
(i) acidic: $[S^{2 -}] = \frac{1.1 \times 1 0 ^{- 21} \times 0.1}{( 0.3 ) ^{2}} = \frac{1.1 \times 1 0 ^{- 22}}{0.09} \approx 1.22 \times 1 0^{- 21}$ M.
(ii) basic: $[S^{2 -}] = \frac{1.1 \times 1 0 ^{- 21} \times 0.1}{( 1 0 ^{- 9} ) ^{2}} = \frac{1.1 \times 1 0 ^{- 22}}{1 0 ^{- 18}} = 1.1 \times 1 0^{- 4}$ M.
Now compare $Q = 0.01 \times [S^{2 -}]$ to $K_{s p}$ :
- $C u S$ ( $K_{s p} \approx 8 \times 1 0^{- 37}$ ): acidic $Q = 0.01 \times 1.22 \times 1 0^{- 21} = 1.22 \times 1 0^{- 23} ≫ K_{s p}$ → precipitates in acid (Group II) ✓.
- $Z n S$ ( $K_{s p} \approx 1.6 \times 1 0^{- 24}$ ): acidic $Q = 1.22 \times 1 0^{- 23}$ ; is that $> K_{s p}$ ? $1.22 \times 1 0^{- 23} > 1.6 \times 1 0^{- 24}$ — yes by ~8×, borderline. Basic $Q = 0.01 \times 1.1 \times 1 0^{- 4} = 1.1 \times 1 0^{- 6} ≫ K_{s p}$ → firmly precipitates in base (Group IV).

Verify: ratio $\frac{[ S ^{2 -} ] _{ba s i c}}{[ S ^{2 -} ] _{a c i d i c}} = \frac{1.1 \times 1 0 ^{- 4}}{1.22 \times 1 0 ^{- 21}} \approx 9 \times 1 0^{16}$ — that's $(0.3/1 0^{- 9})^{2} = (3 \times 1 0^{8})^{2} = 9 \times 1 0^{16}$ ✓, the inverse-square in action. Going basic multiplies sulphide by ~ $1 0^{17}$ , which is why Group IV waits for base.

Cell H — real-world word problem (hardness = Group V logic)

Cell I — the exam twist (common-ion suppression flips the answer)

Worked example Cell I. Two beakers of

$H_{2} S$ -saturated solution both contain $Z n^{2 +} = 0.01$ M. Beaker P has $[H^{+}] = 0.01$ M; beaker Q is the same but a classmate dumps in HCl to make $[H^{+}] = 1$ M. Same $Z n^{2 +}$ , same $H_{2} S$ — yet only one precipitates $Z n S$ . Which, and why? Forecast: does raising $[H^{+}]$ from $0.01$ to $1$ M keep $Z n S$ in or out?

Beaker P: substitute $[H^{+}] = 0.01$ into the master formula, $[S^{2 -}] = \frac{1.1 \times 1 0 ^{- 21} \times 0.1}{( 0.01 ) ^{2}} = \frac{1.1 \times 1 0 ^{- 22}}{1 0 ^{- 4}} = 1.1 \times 1 0^{- 18}$ M. Why this step? We need the actual sulphide level before we can form $Q$ . Then $Q = 0.01 \times 1.1 \times 1 0^{- 18} = 1.1 \times 1 0^{- 20}$ . Compare $K_{s p} (Z n S) = 1.6 \times 1 0^{- 24}$ : $Q ≫ K_{s p}$ → precipitates.
Beaker Q: same substitution but with $[H^{+}] = 1$ : $[S^{2 -}] = \frac{1.1 \times 1 0 ^{- 21} \times 0.1}{( 1 ) ^{2}} = 1.1 \times 1 0^{- 22}$ M. Why redo it? Only $[H^{+}]$ changed, so we re-evaluate the same formula at the new acidity. Then $Q = 0.01 \times 1.1 \times 1 0^{- 22} = 1.1 \times 1 0^{- 24}$ . Compare: $1.1 \times 1 0^{- 24} < 1.6 \times 1 0^{- 24}$ → $Q < K_{s p}$ → stays dissolved.
Why the flip? Extra $H^{+}$ (the Common Ion Effect) pushes both $H_{2} S$ dissociation steps backward (Le Chatelier Principle), starving the solution of $S^{2 -}$ . That is exactly the trick used to keep Group IV ions like $Z n^{2 +}$ in solution during Group II. The two beakers are the two cyan squares straddling the amber threshold in Figure 2.

Verify: raising $[H^{+}]$ from $0.01 \to 1$ (×100) cuts $[S^{2 -}]$ by $10 0^{2} = 1 0^{4}$ , dropping $Q$ below $K_{s p}$ . Same ions, opposite fate — the answer is controlled by acidity, not by what cations are present. That is the whole point of the acidic-vs-basic sulphide split, and it closes our matrix at Cell I.

Recall Rapid self-check

Which direction of $[H^{+}]$ raises $[S^{2 -}]$ ? ::: Lower $[H^{+}]$ (more basic) — because $[S^{2 -}] \propto 1/ [H^{+}]^{2}$ . Why does $P b^{2 +}$ appear in Groups I and II? ::: $P b C l_{2}$ has a moderate $K_{s p}$ ; a residue ( $\sim 1.7 \times 1 0^{- 3}$ M) survives Group I and precipitates as $P b S$ in Group II. At $Q = K_{s p}$ , does solid form? ::: No — that's exact saturation, the knife-edge; you need $Q > K_{s p}$ . Skipping HCl before acidic $H_{2} S$ causes what? ::: Group I sulphides ( $A g_{2} S$ ) co-precipitate with Group II — the I/II separation is lost. Multiplying $[H^{+}]$ by 100 changes $[S^{2 -}]$ by what factor? ::: Divides it by $1 0^{4}$ (inverse-square). Why is $[H_{2} S] \approx 0.1$ M treated as constant? ::: It is the room-temperature saturation level and $H_{2} S$ barely ionises, so it stays put. What do $K_{a 1}$ and $K_{a 2}$ measure? ::: How far each of the two $H_{2} S$ proton-loss steps proceeds; $K_{a 2}$ is tiny, so free $S^{2 -}$ is always scarce.