Solubility product Ksp — common-ion suppression, selective precipitation: Worked examples

This page is the drill hall for the parent topic. The parent note built the tools: $K_{s p}$ , solubility $s$ , the common-ion effect, and the ionic product $Q$ . Here we throw every kind of question at those tools until no scenario can surprise you.

Before a single number: two symbols we lean on the whole page.

The figure below fixes this comparison in your head before we compute a single $Q$ . Read it as a number line drawn along the horizontal axis (labelled "ionic product $Q$ , increasing to the right"): $K_{s p}$ is one fixed dot on that axis; $Q$ is a marker that slides left or right depending on the actual ion concentrations. Everything on this page is deciding which side of the dot $Q$ lands on.

Figure — Solubility product Ksp — common-ion suppression, selective precipitation

The figure is a labelled number line. Its horizontal axis is $Q$ (increasing rightward); the single dark dot is the fixed $K_{s p}$ . The band to the left of the dot is labelled "unsaturated — more can dissolve" ( $Q < K_{s p}$ ); the band to the right is labelled "supersaturated — solid drops out" ( $Q > K_{s p}$ ); the dot itself is labelled "saturated" ( $Q = K_{s p}$ ). (Colours — mint left, coral right, lavender at the dot — merely reinforce these three written labels, so the picture reads the same without relying on colour.) Keep this picture handy: every "does it precipitate?" question is just asking which labelled band $Q$ falls in.

The scenario matrix

Every Ksp problem you will ever meet is one of these cells. The examples below are labelled with the cell they hit, so you can see the whole board is covered.

Cell	Scenario class	What makes it tricky	Example
A	1:1 salt, pure water	baseline, $K_{s p} = s^{2}$	Ex 1
B	1:2 (or 2:1) salt, pure water	a power of 2 appears; watch the $4 s^{3}$	Ex 2
C	common ion added (1:1)	shift, approximation, huge drop	Ex 3
D	common ion, approximation fails	must solve the quadratic honestly	Ex 4
E	mix two solutions → does it precipitate?	dilution on mixing, compute $Q$	Ex 5
F	degenerate / zero input	one ion absent → $Q = 0$ , never precipitates	Ex 6
G	selective precipitation (which drops first?)	two salts compete for one common ion	Ex 7
H	limiting behaviour (as $C \to$ large / small)	check the trend, not one number	Ex 8
I	real-world word problem	hard water / fluoride — read, then model	Ex 9
J	exam twist (pH controls the anion)	anion supply set by acid–base, not a salt	Ex 10

Prerequisites lean on Chemical Equilibrium andrium Constant, Le Chatelier's Principle, Ionic Equilibria in Solutions, Stoichiometry and Solution Concentration, and for Ex 10, pH and pOH Calculations.

Cell A — 1:1 salt in pure water

Worked example Example 1 — the baseline (Cell A)

Problem: $K_{s p}$ of $AgBr$ is $5.0 \times 1 0^{- 13}$ . Find its solubility in pure water.

Forecast: Guess first — will $s$ be near $1 0^{- 6}$ or near $1 0^{- 13}$ ? (Because we square-root a 1:1 salt, expect roughly the half-power, so $\sim 1 0^{- 6}$ – $1 0^{- 7}$ .)

Write the dissolution. $AgBr(s) ⇌ Ag^{+} + Br^{-}$ . Why this step? $K_{s p}$ is the equilibrium constant of this reaction; we cannot use it until the reaction is on paper.
Let $s$ be the solubility. Every formula unit that dissolves gives one $Ag^{+}$ and one $Br^{-}$ , so $[Ag^{+}] = s$ and $[Br^{-}] = s$ . Why this step? The 1:1 stoichiometry ties both ion concentrations to a single unknown $s$ .
Substitute. $K_{s p} = [Ag^{+}] [Br^{-}] = s \cdot s = s^{2}$ , so $5.0 \times 1 0^{- 13} = s^{2}$ . Why this step? At saturation $Q = K_{s p}$ ; that equality is the equation.
Solve. $s = 5.0 \times 1 0^{- 13} = 7.07 \times 1 0^{- 7} mol/L$ .

Verify: Plug back: $(7.07 \times 1 0^{- 7})^{2} = 5.0 \times 1 0^{- 13}$ ✓. And $7 \times 1 0^{- 7}$ landed exactly in the forecast band. Units: $(mol/L)^{2} = mol/L$ ✓.

Cell B — 1:2 salt in pure water

Worked example Example 2 — the sneaky factor of 4 (Cell B)

Problem: $K_{s p}$ of $Ag_{2} CrO_{4}$ is $1.1 \times 1 0^{- 12}$ . Find $s$ in pure water.

Forecast: This salt is 2:1 ( $Ag_{2} CrO_{4} \to 2 Ag^{+} + CrO_{4}^{2 -}$ ). Predict: will $s$ be bigger or smaller than an AgBr-like $1 0^{- 6}$ ? (The cube root of a tiny number is larger than its square root, so expect it a touch bigger, near $1 0^{- 4}$ .)

Dissolution: $Ag_{2} CrO_{4} (s) ⇌ 2 Ag^{+} + CrO_{4}^{2 -}$ .
Stoichiometry: if $s$ dissolves, two silvers appear per unit, so $[Ag^{+}] = 2 s$ and $[CrO_{4}^{2 -}] = s$ . Why this step? The coefficient 2 doubles the silver concentration — miss this and every later number is wrong.
Substitute with correct powers: $K_{s p} = [Ag^{+}]^{2} [CrO_{4}^{2 -}] = (2 s)^{2} (s) = 4 s^{3}$ . Why this step? The exponent on $[Ag^{+}]$ is its coefficient (2), and the factor $2 s$ is its concentration — two separate uses of the same "2".
Solve: $4 s^{3} = 1.1 \times 1 0^{- 12} \Rightarrow s^{3} = 2.75 \times 1 0^{- 13} \Rightarrow s = 6.50 \times 1 0^{- 5} mol/L$ .

Verify: $4 (6.50 \times 1 0^{- 5})^{3} = 4 \times 2.746 \times 1 0^{- 13} = 1.10 \times 1 0^{- 12}$ ✓. Forecast said "near $1 0^{- 4}$ " — $6.5 \times 1 0^{- 5}$ is right there ✓.

Cell C — common ion, approximation holds

Worked example Example 3 — common ion suppresses (Cell C)

Problem: Solubility of $AgBr$ ( $K_{s p} = 5.0 \times 1 0^{- 13}$ ) in $0.10 M$ $KBr$ .

Forecast: KBr floods the solution with $Br^{-}$ . By Le Chatelier, the AgBr equilibrium shifts left. Predict: much less or much more soluble than pure water's $7 \times 1 0^{- 7}$ ?

Existing $[Br^{-}]$ : KBr is fully soluble, so $[Br^{-}]_{start} = 0.10 M$ .
Let $s^{'}$ = new solubility. Then $[Ag^{+}] = s^{'}$ and $[Br^{-}] = 0.10 + s^{'}$ . Why this step? The bromide now comes from two sources: 0.10 from KBr, plus $s^{'}$ from the tiny bit of AgBr that dissolves.
Approximate: since AgBr barely dissolves, guess $s^{'} ≪ 0.10$ , so $[Br^{-}] \approx 0.10$ . Why this step? It turns a quadratic into one-line algebra — provided we check the guess at the end.
Solve: $5.0 \times 1 0^{- 13} = s^{'} (0.10) \Rightarrow s^{'} = 5.0 \times 1 0^{- 12} mol/L$ .

Verify: Check the approximation: $5.0 \times 1 0^{- 12} ≪ 0.10$ ✓ (a factor $2 \times 1 0^{10}$ ). Compare to pure water: $\frac{7.07 \times 1 0 ^{- 7}}{5.0 \times 1 0 ^{- 12}} \approx 1.4 \times 1 0^{5}$ — ~140 000× less soluble. Forecast "much less" ✓.

Cell D — common ion, approximation FAILS

Worked example Example 4 — when you must face the quadratic (Cell D)

Problem: A hypothetical salt $MX$ has $K_{s p} = 2.0 \times 1 0^{- 3}$ — chosen deliberately large so the approximation must break (real "slightly insoluble" salts such as $PbCl_{2}$ , $K_{s p} \approx 1.6 \times 1 0^{- 5}$ , or $CaSO_{4}$ , $K_{s p} \approx 2.4 \times 1 0^{- 5}$ , are smaller, but the same quadratic method applies whenever $s^{'}$ is not $≪ C$ ). Find $s^{'}$ in $0.010 M$ $NaX$ .

Forecast: Here $K_{s p}$ is only ~5× smaller than the common-ion concentration squared. Will $s^{'} ≪ 0.010$ still hold? Guess no — and that's the whole point of this cell.

Setup: $[M^{+}] = s^{'}$ , $[X^{-}] = 0.010 + s^{'}$ , so $s^{'} (0.010 + s^{'}) = 2.0 \times 1 0^{- 3}$ .
Naive attempt (to expose the trap): drop $s^{'}$ beside 0.010 → $s^{'} = \frac{2.0 \times 1 0 ^{- 3}}{0.010} = 0.20 M$ . Why this step? We test the shortcut. But $0.20 \neq ≪ 0.010$ — it's 20× bigger. The approximation is invalid; we cannot use it.
Solve the true quadratic. Expand: $s^{'2} + 0.010 s^{'} - 2.0 \times 1 0^{- 3} = 0$ . Why this step? No shortcut survives, so we keep the full $(0.010 + s^{'})$ term and use $s^{'} = \frac{- b + b ^{2} - 4 a c}{2 a}$ with $a = 1, b = 0.010, c = - 2.0 \times 1 0^{- 3}$ .
Compute: discriminant $= 0.01 0^{2} + 4 (2.0 \times 1 0^{- 3}) = 1.0 \times 1 0^{- 4} + 8.0 \times 1 0^{- 3} = 8.1 \times 1 0^{- 3}$ ; $= 0.09000$ . $s^{'} = \frac{- 0.010 + 0.09000}{2} = \frac{0.08000}{2} = 0.04000 mol/L .$

Verify: Plug back: $[X^{-}] = 0.010 + 0.040 = 0.050$ ; $Q = 0.040 \times 0.050 = 2.0 \times 1 0^{- 3} = K_{s p}$ ✓. Note $s^{'} = 0.040$ , not the naive $0.20$ — the shortcut would have been 5× too big. Lesson: always test $s^{'} ≪ C$ ; if it fails, solve the quadratic.

Cell E — mix two solutions, does it precipitate?

Worked example Example 5 — dilution-on-mixing (Cell E)

Problem: Mix $100 mL$ of $0.010 M$ $AgNO_{3}$ with $100 mL$ of $0.020 M$ $NaCl$ . Does AgCl ( $K_{s p} = 1.8 \times 1 0^{- 10}$ ) precipitate?

Forecast: We must compare $Q$ to $K_{s p}$ . The classic slip is to forget that mixing dilutes both solutions. Predict: after halving both concentrations, is $Q$ still above $1.8 \times 1 0^{- 10}$ ?

Total volume after mixing $= 100 + 100 = 200 mL = 0.200 L$ . Why this step? Concentration is moles ÷ volume; the volume just doubled, so every concentration drops.
Dilute each ion (moles conserved, new volume 0.200 L): $[\text{Cl}^-]=\frac{0.020\times0.100}{0.200}=1.0\times10^{-2}\ \text{M}.$$ *Why this step?* Only the **diluted** concentrations describe the actual mixed beaker.$
Compute $Q$ : $Q = [Ag^{+}] [Cl^{-}] = (5.0 \times 1 0^{- 3}) (1.0 \times 1 0^{- 2}) = 5.0 \times 1 0^{- 5}$ . Why this step? $Q$ is our "live" marker on the number line of the opening figure; we build it from the actual diluted concentrations so we can see which labelled band it lands in.
Compare: $Q = 5.0 \times 1 0^{- 5}$ vs $K_{s p} = 1.8 \times 1 0^{- 10}$ . Since $Q ≫ K_{s p}$ (about $2.8 \times 1 0^{5}$ times bigger), precipitation occurs. Why this step? The comparison is the decision rule: $Q$ sits deep in the supersaturated (right-hand) band, so solid must crash out until $Q$ falls back to $K_{s p}$ .

Verify: Sanity — both reactants are far above the $\sim 1 0^{- 5}$ saturation level found for AgCl in the parent note, so a solid must appear; $Q > K_{s p}$ agrees ✓. Units of $Q$ : $(mol/L)^{2}$ ✓.

Cell F — degenerate / zero input

Worked example Example 6 — one ion is missing (Cell F)

Problem: You have $0.10 M$ $AgNO_{3}$ but zero chloride ( $[Cl^{-}] = 0$ ). Can AgCl precipitate? What $[Cl^{-}]$ is needed to just begin precipitation?

Forecast: With no chloride at all, guess whether any solid can form — and if not, at what chloride level it starts.

Current $Q$ : $Q = [Ag^{+}] [Cl^{-}] = (0.10) (0) = 0$ . Why this step? Any product with a zero factor is zero — this is the degenerate case.
Compare: $Q = 0 < K_{s p} = 1.8 \times 1 0^{- 10}$ . Since $Q < K_{s p}$ , the solution is unsaturated: no precipitate, ever, while $[Cl^{-}] = 0$ . Why this step? Precipitation needs $Q \geq K_{s p}$ ; a zero product can't reach the ceiling.
Onset condition: precipitation begins the instant $Q = K_{s p}$ : $[Cl^{-}]_{onset} = \frac{K _{s p}}{[ Ag ^{+} ]} = \frac{1.8 \times 1 0 ^{- 10}}{0.10} = 1.8 \times 1 0^{- 9} M .$ Why this step? Solve $Q = K_{s p}$ for the missing ion to find the threshold.

Verify: At onset, $Q = (0.10) (1.8 \times 1 0^{- 9}) = 1.8 \times 1 0^{- 10} = K_{s p}$ ✓. The threshold is astonishingly tiny — even a whiff of $Cl^{-}$ clouds a silver solution, which is exactly why the $Cl^{-}$ test in Qualitative Inorganic Analysis is so sensitive.

Cell G — selective precipitation

Worked example Example 7 — which salt drops first? (Cell G)

Problem: A solution is $0.10 M$ in $Cl^{-}$ and $0.10 M$ in $CrO_{4}^{2 -}$ . You add $Ag^{+}$ slowly. Given $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ and $K_{s p} (Ag_{2} CrO_{4}) = 1.1 \times 1 0^{- 12}$ , which precipitates first, and how well are they separated?

Forecast: Both compete for $Ag^{+}$ . The one needing the lowest $[Ag^{+}]$ to reach its $K_{s p}$ precipitates first. Guess: AgCl or $Ag_{2} CrO_{4}$ ?

Silver needed for AgCl: $[Ag^{+}]_{1} = \frac{K _{s p}}{[ Cl ^{-} ]} = \frac{1.8 \times 1 0 ^{- 10}}{0.10} = 1.8 \times 1 0^{- 9} M$ . Why this step? Precipitation of AgCl starts exactly when $Q = K_{s p}$ ; solving that equality for $[Ag^{+}]$ gives the silver level that first "fills the ceiling" for chloride.
Silver needed for $Ag_{2} CrO_{4}$ : here $K_{s p} = [Ag^{+}]^{2} [CrO_{4}^{2 -}]$ , so $[Ag^{+}]_{2} = \frac{K _{s p}}{[ CrO _{4}^{2 -} ]} = \frac{1.1 \times 1 0 ^{- 12}}{0.10} = 1.1 \times 1 0^{- 11} = 3.32 \times 1 0^{- 6} M .$ Why this step? The 2:1 salt needs $[Ag^{+}]$ squared, so we square-root — a different algebra from the 1:1 case.
Compare thresholds: $1.8 \times 1 0^{- 9} < 3.32 \times 1 0^{- 6}$ , so AgCl precipitates first (it reaches its ceiling at ~1800× lower silver). Why this step? As we drip silver in, $[Ag^{+}]$ rises past the smaller threshold first; whichever salt's threshold is crossed first is the one that precipitates first.
Purity of separation: find $[Cl^{-}]$ remaining when chromate is about to drop, i.e. when $[Ag^{+}] = 3.32 \times 1 0^{- 6}$ : $[Cl^{-}]_{left} = \frac{K _{s p} ( AgCl )}{[ Ag ^{+} ]} = \frac{1.8 \times 1 0 ^{- 10}}{3.32 \times 1 0 ^{- 6}} = 5.4 \times 1 0^{- 5} M .$ Why this step? This tells us how much chloride is still in solution when chromate begins — the leftover fraction measures separation quality.

Verify: Fraction of chloride still dissolved $= \frac{5.4 \times 1 0 ^{- 5}}{0.10} = 5.4 \times 1 0^{- 4} = 0.054%$ . So >99.9% of the chloride is already solid before the first chromate falls — an excellent selective precipitation, the principle behind Mohr's titration ✓.

The figure below plots both silver thresholds on a log axis so you can see the ~1800× gap that makes the separation clean — AgCl's threshold line sits far to the left of chromate's line.

Read the picture: the horizontal axis is added $[Ag^{+}]$ on a log scale (increasing rightward). Two vertical threshold lines are labelled — "AgCl needs $1.8 \times 1 0^{- 9}$ M" (left) and " $Ag_{2} CrO_{4}$ needs $3.3 \times 1 0^{- 6}$ M" (right). As you add silver (move rightward), you cross the AgCl line long before the chromate line. In the wide gap between them, essentially all chloride has already precipitated while chromate is still fully dissolved — that gap is the separation window, and its width is precisely why selective precipitation works. The lesson to carry away: when two salts share a common ion, order and quality of separation are decided entirely by comparing the ion concentration each one needs to hit its own $K_{s p}$ .

Cell H — limiting behaviour

Worked example Example 8 — the trend, not one number (Cell H)

Problem: For AgCl ( $K_{s p} = 1.8 \times 1 0^{- 10}$ ), how does solubility $s^{'} \approx K_{s p} / C$ behave as the common-ion concentration $C$ goes from very small to very large? Evaluate at $C = 1 0^{- 4}, 1 0^{- 2}, 1 M$ and describe the limits.

Forecast: Predict the shape: as $C$ grows 100×, does $s^{'}$ shrink 100×, or something gentler?

Model: $s^{'} \approx \frac{K _{s p}}{C}$ (valid while $s^{'} ≪ C$ ).
Evaluate: $C=10^{-2}:\ s'=1.8\times10^{-8};\quad C=1:\ s'=1.8\times10^{-10}.$$ *Why this step?* Three points reveal the law: each 100× rise in $C$ gives a 100× fall in $s'$ — perfectly inverse.$
Small- $C$ limit: as $C \to 0$ , the formula $s^{'} = K_{s p} / C$ diverges — but it is only valid while $C ≫ s^{'}$ . Below that point the common ion no longer dominates and we fall back to the pure-water case, where $s \to K_{s p} = 1.34 \times 1 0^{- 5}$ . Interpretation: solubility does not blow up to infinity; it is capped at the pure-water value. This is the answer to the forecast's implicit worry — the trend is inverse only in the valid range, then it flattens to a finite ceiling. Why this step? A limit is meaningless if the model that produces it has already broken; honest limiting analysis states where the formula stops applying.
Large- $C$ limit: as $C \to \infty$ , $s^{'} = K_{s p} / C \to 0$ . Interpretation: flooding with common ion drives solubility asymptotically toward zero — the strongest possible suppression, and the whole reason common-ion washing is used to recover a precipitate almost quantitatively. The two limits together bracket the behaviour: bounded above by $K_{s p}$ at tiny $C$ , squeezed toward $0$ at huge $C$ . Why this step? Naming both endpoints turns three isolated numbers into a complete story of the curve.

Verify: On log–log axes, $s^{'}$ vs $C$ is a straight line of slope $- 1$ (inverse proportionality) in the valid range ✓; product $s^{'} \cdot C = K_{s p} = 1.8 \times 1 0^{- 10}$ at all three points ✓; and the small- $C$ cap $1.8 \times 1 0^{- 10} = 1.34 \times 1 0^{- 5}$ matches the pure-water solubility of AgCl ✓.

Cell I — real-world word problem

Worked example Example 9 — fluoridated water (Cell I)

Problem: Tap water is fluoridated to $[F^{-}] = 1.0 \times 1 0^{- 4} M$ (a typical $\approx 1.9 ppm$ ). Ground water carries $[Ca^{2 +}] = 2.0 \times 1 0^{- 3} M$ . For $CaF_{2}$ , $K_{s p} = 3.9 \times 1 0^{- 11}$ . Will $CaF_{2}$ scale form in the pipes?

Forecast: Read → model → compute $Q$ . Guess: with fluoride that dilute, does $Q$ clear the $K_{s p}$ bar?

Translate to ions. The salt is $CaF_{2}$ , so (recalling $Q = [cation]^{x} [anion]^{y}$ ) $Q = [Ca^{2 +}] [F^{-}]^{2}$ — fluoride is squared (two F per formula). Why this step? The word problem hides the stoichiometry; the coefficient 2 must resurface as an exponent inside $Q$ .
Compute $Q$ : $Q = (2.0 \times 1 0^{- 3}) (1.0 \times 1 0^{- 4})^{2} = (2.0 \times 1 0^{- 3}) (1.0 \times 1 0^{- 8}) = 2.0 \times 1 0^{- 11} .$ Why this step? $Q$ is the live marker we always build first; only after we have a number can we compare it to the fixed $K_{s p}$ ceiling.
Compare: $Q = 2.0 \times 1 0^{- 11}$ vs $K_{s p} = 3.9 \times 1 0^{- 11}$ . Since $Q < K_{s p}$ , the water is unsaturated — no scale forms. Why this step? $Q < K_{s p}$ (the left-hand, unsaturated band of the opening figure) means room to dissolve, not precipitate.

Verify: $Q / K_{s p} = 2.0/3.9 = 0.51$ — about half the ceiling, safely below ✓. Real check: fluoridation levels are set precisely to stay under $CaF_{2}$ saturation, which agrees ✓.

Cell J — exam twist: pH controls the anion

Worked example Example 10 — the anion is a base (Cell J)

Problem: $Mg(OH)_{2}$ has $K_{s p} = 5.6 \times 1 0^{- 12}$ . A buffered solution holds $pH = 9.0$ . Will $Mg(OH)_{2}$ precipitate from $0.010 M$ $Mg^{2 +}$ ?

Forecast: The twist: $[OH^{-}]$ is fixed by pH, not by adding a hydroxide salt. You'll need pH and pOH Calculations. Guess whether pH 9 supplies enough hydroxide to precipitate.

Get $[OH^{-}]$ from pH. $pOH = 14 - 9 = 5$ , so $[OH^{-}] = 1 0^{- 5} M$ . Why this step? Hydroxide is the common/participating ion here, and its supply is set by acid–base equilibrium, not a dissolved salt — so we must extract it from pH first.
Write $Q$ with the correct power. $Mg(OH)_{2} ⇌ Mg^{2 +} + 2 OH^{-}$ , so $Q = [Mg^{2 +}] [OH^{-}]^{2}$ . Why this step? Two hydroxides per formula unit means $[OH^{-}]$ enters $Q$ squared; this is the same "coefficient becomes exponent" rule as every other cell.
Compute: $Q = (0.010) (1 0^{- 5})^{2} = (1 0^{- 2}) (1 0^{- 10}) = 1.0 \times 1 0^{- 12}$ . Why this step? Build the live $Q$ from the actual $[Mg^{2 +}]$ and the pH-derived $[OH^{-}]$ so we can compare to $K_{s p}$ .
Compare: $Q = 1.0 \times 1 0^{- 12} < K_{s p} = 5.6 \times 1 0^{- 12}$ . Since $Q < K_{s p}$ , no precipitate at pH 9. Why this step? $Q$ sits in the unsaturated (left-hand) band — the pH is too low (not basic enough) to push $Q$ over the ceiling.
Bonus — onset pH: precipitation begins when $Q = K_{s p}$ : $[OH^{-}] = \frac{K _{s p}}{[ Mg ^{2 +} ]} = \frac{5.6 \times 1 0 ^{- 12}}{0.010} = 5.6 \times 1 0^{- 10} = 2.37 \times 1 0^{- 5} M,$ giving $pOH = - lo g (2.37 \times 1 0^{- 5}) = 4.63$ , so $pH_{onset} = 14 - 4.63 = 9.37$ . Why this step? Solving $Q = K_{s p}$ for $[OH^{-}]$ and converting back to pH tells us the exact pH where the unsaturated band ends and precipitation begins.

Verify: At pH 9.37, $[OH^{-}] = 2.37 \times 1 0^{- 5}$ and $Q = (0.010) (2.37 \times 1 0^{- 5})^{2} = 5.6 \times 1 0^{- 12} = K_{s p}$ ✓. Since our pH 9.0 < 9.37, no solid — consistent with step 4 ✓. This is exactly how Le Chatelier's Principle plus pH control hydroxide precipitations in Qualitative Inorganic Analysis.

Retrieval

Recall Which comparison decides whether a precipitate forms?

Compare $Q$ (live ionic product) to $K_{s p}$ (fixed ceiling) ::: $Q > K_{s p}$ precipitates, $Q = K_{s p}$ saturated, $Q < K_{s p}$ unsaturated.

Recall For a 2:1 salt like

$Ag_{2} CrO_{4}$ , why is $K_{s p} = 4 s^{3}$ and not $s^{3}$ ? Because $[Ag^{+}] = 2 s$ enters squared: $(2 s)^{2} \cdot s = 4 s^{3}$ ::: the coefficient 2 acts once as a factor in the concentration and once as the exponent.

Recall When you mix two solutions before computing

$Q$ , what must you do first? Dilute every concentration to the new total volume (moles conserved) ::: only then compute $Q$ .

Recall When does the common-ion shortcut

$s^{'} \approx K_{s p} / C$ fail? When $s^{'}$ is not $≪ C$ (typically when $K_{s p}$ is large or $C$ is small) ::: then solve the full quadratic.

Recall In selective precipitation, which salt precipitates first?

The one requiring the lowest concentration of the added common ion to reach its $K_{s p}$ ::: check each threshold and pick the smaller.