Question bank — Cation groups I–V — group reagents, separation scheme

False — because $K_{a2}\sim 10^{-13}$ is minuscule, almost all dissolved sulphide sits as $H{S}^{-}$ and undissociated $H_{2}S$; $[S^{2-}]$ is only a tiny derived tail, and that smallness is what makes Group II selective. See Weak Acid Dissociation — H2S.

False — more base does raise $[S^{2-}]$ and $[O{H}^{-}]$, but that destroys selectivity: Group IV ions crash out with Group II, and you can no longer tell the groups apart.

True — $PbC{l}_2$ has a moderate (borderline) $K_{sp}$ and is partly soluble, especially in hot water, so most $P{b}^{2+}$ leaves in Group I but a dissolved residue carries over and reappears as $PbS$ in Group II.

False — it is the reverse: the most selective / least aggressive reagent (dil. HCl) goes first, then progressively less picky ones, so each stage removes only its own group without stealing later ones.

False — the solution is kept roughly saturated ($\sim 0.1$ M) and $H_{2}S$ is barely ionised, so $[H_{2}S]$ stays nearly constant; what changes is $[H^{+}]$, which controls $[S^{2-}]$ through $[S^{2-}]=K_{a1}{K}_{a2}[H_{2}S]/[H^{+}{]}^2$.

False — the $H^{+}$ from HCl suppresses $H_{2}S$ ionisation by the Common Ion Effect, keeping $[S^{2-}]$ low so that Group IV sulphides stay dissolved and don't contaminate Group II.

False — precipitation depends on the actual $K_{sp}$ value and on the cation concentration in the ionic product $Q$; a smaller $K_{sp}$ salt precipitates at a lower anion level. See Solubility Product Ksp.

True — $M{g}^{2+},N{a}^{+},K^{+},N{H}_4^{+}$ don't precipitate with any of the five reagents under the standard conditions, so they stay in solution and are identified by flame or special tests.

The error is confusing maximising precipitate with achieving separation; high $[S^{2-}]$ precipitates Group IV too. In Group II we deliberately keep $[S^{2-}]$ low (acidic) so only the tiniest-$K_{sp}$ sulphides pass the $Q>K_{sp}$ threshold.

Without $N{H}_{4}Cl$, $[O{H}^{-}]$ climbs too high and Group IV hydroxides like $Zn(OH{)}_2$ also precipitate. $N{H}_{4}Cl$ supplies $N{H}_4^{+}$ that suppresses $O{H}^{-}$ (Common Ion Effect), holding it just high enough for the very-insoluble Group III hydroxides only.

Hitting many cations is not strength but lack of selectivity. Starting with $H_{2}S$ would merge Groups I and II (in acid) and destroy the whole ordered separation; always go I → II → III → IV → V.

The precise onset is $Q>K_{sp}$; at $Q=K_{sp}$ the solution is exactly saturated (equilibrium), and any further increase in ionic product tips it into forming solid.

More $H_{2}S$ barely helps because $[H_{2}S]$ is already saturated and $[S^{2-}]\propto 1/[H^{+}{]}^2$; the limiting factor is $[H^{+}]$. You must lower $[H^{+}]$ (go basic with $N{H}_{4}OH$), not add more gas.

Only cations whose chlorides have very small $K_{sp}$ ($A{g}^{+},P{b}^{2+},H{g}_2^{2+}$) precipitate; many other chlorides are soluble, so forming a chloride on paper is not enough — the ionic product must exceed a tiny $K_{sp}$.

Because their sulphides span a huge range of $K_{sp}$; by tuning pH we tune $[S^{2-}]$, so the tiny-$K_{sp}$ ones (II) precipitate in acid at low $[S^{2-}]$ while the larger-$K_{sp}$ ones (IV) need the high $[S^{2-}]$ of basic medium.

Because sulphide is released in two protonation-reversal steps ($H_{2}S\to H{S}^{-}\to S^{2-}$); each free $S^{2-}$ requires losing two $H^{+}$, so combining $K_{a1}$ and $K_{a2}$ puts $[H^{+}{]}^2$ in the denominator.

The basic $N{H}_{4}OH$ medium needed for Group IV would also precipitate the Group III hydroxides; removing $Fe(OH{)}_3,Al(OH{)}_3,Cr(OH{)}_3$ first keeps the Group IV sulphide precipitate clean and unambiguous.

$N{H}_4^{+}$ is a common ion of the equilibrium $N{H}_{4}OH\rightleftharpoons N{H}_4^{+}+O{H}^{-}$; by Le Chatelier Principle extra $N{H}_4^{+}$ shifts it left, cutting $[O{H}^{-}]$ so only the smallest-$K_{sp}$ hydroxides precipitate.

We never add $S^{2-}$ directly; we control it indirectly by choosing pH, since $[S^{2-}]=K_{a1}{K}_{a2}[H_{2}S]/[H^{+}{]}^2$ — the sulphide level is an output of the equilibrium, tunable through $[H^{+}]$.

Colour and single-test reactions overlap for many ions, but $K_{sp}$ values differ enough that by controlling one anion's concentration we can force only selected cations past $Q>K_{sp}$, giving a reproducible fractional separation.

Each reagent is less selective than the last; if a later, more aggressive reagent is used early, it precipitates ions belonging to earlier groups too, collapsing distinct groups together and ruining identification.

$PbC{l}_2$ is appreciably soluble in hot water, so more $P{b}^{2+}$ dissolves back and is carried into Group II — this is exactly why lead can be confirmed in two places, so the borderline case is expected.

Nothing precipitates in Groups I–IV because none of these ions forms a low-$K_{sp}$ chloride, sulphide, or hydroxide under the conditions; $C{a}^{2+}$ appears in Group V as $CaC{O}_3$, while $N{a}^{+}$ and $K^{+}$ remain and are found by flame tests.

No — the ionic product $Q=[M^{2+}][S^{2-}]$ can fall below $K_{sp}$ if $[M^{2+}]$ is tiny, so precipitation may not occur even for a genuine Group II ion; the group is defined by the $K_{sp}$ of its sulphide, not guaranteed to precipitate at any concentration.

$[S^{2-}]=K_{a1}{K}_{a2}[H_{2}S]/[H^{+}{]}^2$ tends toward zero as $[H^{+}]\to$ large, which is why only the very-smallest-$K_{sp}$ sulphides (Group II) can still satisfy $Q>K_{sp}$.

$[S^{2-}]$ grows very large as $[H^{+}]\to$ small, so even the larger-$K_{sp}$ Group IV sulphides now satisfy $Q>K_{sp}$ and precipitate — this is the basis of the basic-medium Group IV step.

Group I chlorides never separate, and Group I ions that also form insoluble sulphides come down mixed with Group II, so two groups are contaminated together and cannot be distinguished. See Confirmatory Tests for Cations.