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ChemistryIonic Equilibrium / Solubility

Arrange the following in decreasing order of solubility in water:
$\text{Zn(OH)}_2\ (K_{sp} = 3 \times 10^{-17})$, $\text{AgBr}\ (K_{sp} = 5 \times 10^{-13})$, $\text{Hg}_2\text{Cl}_2\ (K_{sp} = 1.4 \times 10^{-18})$

R
Solution written and verified by Roshan, science educator with 5 years of experience teaching NEET and JEE aspirants. Last reviewed September 2026.
Options
1
AgBr > Zn(OH)$_2$ > Hg$_2$Cl$_2$
2
Zn(OH)$_2$ > AgBr > Hg$_2$Cl$_2$
3
Hg$_2$Cl$_2$ > AgBr > Zn(OH)$_2$
4
Zn(OH)$_2$ > Hg$_2$Cl$_2$ > AgBr
Correct Answer
Zn(OH)$_2$ > AgBr > Hg$_2$Cl$_2$
Solution
1

Zn(OH)2: $s = (3\times10^{-17}/4)^{1/3} \approx 1.96\times10^{-6}$ M

AgBr: $s = \sqrt{5\times10^{-13}} \approx 7.1\times10^{-7}$ M

2

Hg2Cl2: $s = (1.4\times10^{-18}/4)^{1/3} \approx 7.0\times10^{-7}$ M

Order: $1.96\times10^{-6} > 7.1\times10^{-7} > 7.0\times10^{-7}$

Answer: Zn(OH)2 > AgBr > Hg2Cl2

Calculate s from Ksp for each type: MA2: s=(Ksp/4)^(1/3); MA: s=sqrt(Ksp)
Zn(OH)2 most soluble despite lowest Ksp? No: Zn(OH)2 Ksp=3e-17 > Hg2Cl2 1.4e-18
Theory: Ionic Equilibrium / Solubility
1. Solubility Product (Ksp) — Concept and Derivation

The solubility product constant (Ksp) is the equilibrium constant for the dissolution equilibrium of a sparingly soluble ionic compound in water. For a generic salt MxAy that dissolves according to: MxAy(s) ⇌ xM^(y+)(aq) + yA^(x-)(aq), the equilibrium expression is: Ksp = [M^(y+)]^x × [A^(x-)]^y. The activity of the undissolved solid (pure MxAy) is 1 by convention (just as the activity of pure liquid water is 1 in the Kw expression). Note that Ksp does not include the concentration of the solid — only the dissolved ion concentrations appear. The molar solubility s (in mol/L) of a sparingly soluble salt is related to Ksp by substituting equilibrium concentrations in terms of s: for MxAy: [M^(y+)] = xs and [A^(x-)] = ys. Therefore: Ksp = (xs)^x × (ys)^y = x^x × y^y × s^(x+y). Solving for s: s = [Ksp/(x^x × y^y)]^(1/(x+y)). The key insight: salts with the same formula type (same x and y, and therefore the same Ksp expression in terms of s) have the same Ksp-solubility relationship and can be directly compared by their Ksp values. But salts with different formula types CANNOT be directly compared by Ksp alone — you must calculate the actual molar solubility s for each and then compare.

2. Solubility and Factors Affecting It

The solubility of ionic compounds in water depends on multiple factors that can either increase or decrease the concentration of ions in solution at equilibrium. Temperature: for most ionic compounds, solubility increases with temperature (endothermic dissolution). Examples: KNO3 (dramatically more soluble at higher T), NaCl (slightly increases). Some salts have retrograde solubility (less soluble at higher T): CaSO4, Ca(OH)2, Li2SO4 (these have exothermic dissolution enthalpies). This retrograde behaviour of CaSO4 causes scaling in hot water pipes and heat exchangers. Nature of solute and solvent: "like dissolves like" — polar ionic compounds dissolve well in polar solvents like water; non-polar compounds dissolve well in non-polar solvents. The dissolution of ionic compounds in water is driven by two factors: the lattice energy (must be overcome) and the hydration energy of the ions (released as ions are solvated by water dipoles). If hydration energy > lattice energy: dissolves readily (like NaOH, LiCl). If lattice energy >> hydration energy: low solubility (like AgCl, BaSO4). Lattice energy correlates with ion charges and inversely with ion size. Highly charged, small ions have very high lattice energies: Mg2+, Ca2+, Al3+, Fe3+ salts are generally less soluble than corresponding Na+ or K+ salts for the same anion. pH effects (discussed earlier). Complex ion formation: forming a stable complex with a ligand increases solubility (because complex formation effectively removes the free metal ion from solution, driving the dissolution equilibrium forward). Adding CN- dissolves AgCl (forms [Ag(CN)2]-); adding NH3 dissolves AgCl (forms [Ag(NH3)2]+).

3. Qualitative Analysis Using Selective Precipitation

Selective precipitation exploits differences in Ksp values to separate and identify different cations in a mixture by adding anions that precipitate one cation at a lower concentration than required to precipitate another. This is the basis of the classical qualitative analysis scheme (systematic qualitative analysis, developed primarily by H. Rose and C. R. Fresenius in the 19th century): Group I (chloride group): add dilute HCl → precipitate AgCl (white), PbCl2 (white), Hg2Cl2 (white). Other metal chlorides remain soluble. Group II (acid sulfide group): in acidic solution (0.3 M HCl), add H2S → precipitate CuS (black), PbS (black), CdS (yellow), Bi2S3 (black), As2S3 (yellow), Sb2S3 (orange), SnS2 (yellow). The acidic conditions keep [S2-] low (suppresses HS- dissociation), so only the least soluble sulfides (lowest Ksp) precipitate. Group III (alkaline sulfide and hydroxide group): in alkaline solution (NH4OH/NH4Cl buffer), add H2S → precipitate CoS (black), NiS (black), MnS (pink), FeS (black), ZnS (white), Al(OH)3 (white), Cr(OH)3 (grey-green). Group IV (carbonate group): add (NH4)2CO3 → precipitate BaCO3 (white), SrCO3 (white), CaCO3 (white). Group V (soluble group): Mg2+, Na+, K+, NH4+ remain in solution, confirmed by specific flame tests and individual reactions. The logic of each group separation depends on the Ksp differences: by carefully controlling [S2-] through pH adjustment, one can selectively precipitate Cu2+ (Ksp CuS = 6×10^-36) but not Mn2+ (Ksp MnS = 2.5×10^-13) in the same solution — a difference of 22 orders of magnitude in Ksp.

4. Ionic Equilibria — Buffer Solutions and Henderson-Hasselbalch

Buffer solutions are solutions that resist changes in pH upon addition of small amounts of strong acid or strong base. They consist of: weak acid + its conjugate base (acid buffer, pH < 7 at typical concentrations): e.g., CH3COOH + CH3COO-Na+, H2CO3 + NaHCO3, H2PO4- + HPO4^2-. Weak base + its conjugate acid (basic buffer, pH > 7): e.g., NH3 + NH4Cl. Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) (for acid buffer). This equation shows that: pH = pKa when [A-] = [HA] (half-equivalence point in a titration). Buffer capacity is highest when [A-] = [HA] (pH = pKa) — the buffer can neutralise equal amounts of added acid and base. Buffer range: pH = pKa ± 1 (within this range, both acid and conjugate base are present in significant amounts to provide buffering). Buffer mechanism: added H+ is neutralised by A-: H+ + A- → HA. Added OH- is neutralised by HA: OH- + HA → A- + H2O. As long as [A-] and [HA] are present in significant quantities, the pH changes minimally. Blood buffer system: the most important physiological buffer is the H2CO3/HCO3- system (pKa = 6.1), maintained at [HCO3-]/[H2CO3] = 20/1 to give blood pH ≈ 7.4. The respiratory system adjusts [CO2] (and therefore [H2CO3]) through ventilation rate; the kidneys adjust [HCO3-] through renal excretion. This bicarbonate buffer system, together with plasma protein and haemoglobin buffers, maintains blood pH within the very narrow range 7.35-7.45 that is essential for normal enzyme function.

5. Ionic Product of Water and pH Scale

The self-ionisation of water (also called autoprotolysis or autoionisation) is a crucial equilibrium for understanding all aqueous chemistry: H2O ⇌ H+ + OH- (or more precisely: 2H2O ⇌ H3O+ + OH-). The equilibrium constant for this process is the ionic product of water: Kw = [H+][OH-] = [H3O+][OH-] = 1.0 × 10^-14 at 25°C. Since pure water is neutral, [H+] = [OH-] = sqrt(Kw) = 1.0 × 10^-7 mol/L. The pH scale: pH = -log[H+] (or more rigorously, -log(a_H+) where a_H+ is the hydrogen ion activity). At 25°C in pure water: pH = pOH = 7.0 (neutral). pH < 7: acidic ([H+] > [OH-]). pH > 7: alkaline/basic ([OH-] > [H+]). pH + pOH = pKw = 14 at 25°C. Kw is temperature-dependent: at higher temperatures, Kw increases (ionisation is endothermic), so the neutral pH decreases: at 37°C (body temperature), Kw = 2.4 × 10^-14, neutral pH = 6.81. At 0°C, Kw = 1.1 × 10^-15, neutral pH = 7.47. The pH scale: 0 (1 M HCl, extremely acidic) → 7 (neutral water at 25°C) → 14 (1 M NaOH, extremely basic). In practice, pH values outside 0-14 are possible (negative pH for very concentrated strong acids; pH > 14 for very concentrated strong bases), but the scale is most useful in the range 0-14 for typical biological and chemical systems.

6. Applications of Solubility and Ionic Equilibria in Medicine and Industry

The principles of solubility product, common ion effect, complex formation, and pH-dependent solubility have extensive practical applications in medicine, environmental science, and industry. Renal stones (kidney stones): the most common types are calcium oxalate (CaC2O4, Ksp = 2.3×10^-9) and calcium phosphate (Ca3(PO4)2, Ksp = 2.1×10^-33). Formation occurs when the concentrations of Ca2+ and oxalate/phosphate ions exceed their respective Ksp values. Risk factors: low urine volume (concentrated urine), high dietary oxalate, hypercalciuria. Prevention: high fluid intake (dilutes urine), citrate supplementation (citrate forms soluble complex with Ca2+ and also inhibits CaC2O4 crystal growth), dietary modifications. Treatment: for urate stones (from uric acid, relatively pH-dependent solubility), alkalinisation of urine with sodium bicarbonate increases uric acid solubility; for calcium oxalate and phosphate stones, extracorporeal shock wave lithotripsy (ESWL) or surgery. Dental enamel: composed primarily of hydroxyapatite Ca10(PO4)6(OH)2. Dental caries (cavities) result from acid dissolution of enamel: H+ (from bacterial fermentation of sugars) + Ca10(PO4)6(OH)2 → dissolution of hydroxyapatite. Fluoridation of water (at 0.7-1.0 ppm F-) converts some hydroxyapatite to fluorapatite Ca10(PO4)6F2, which has lower Ksp and is more resistant to acid dissolution. Toothpaste contains fluoride (NaF or stannous fluoride) for the same reason. Wastewater treatment: heavy metal ions (Pb2+, Cd2+, Hg2+, Cu2+, Ni2+, Cr3+) are removed from industrial wastewater by precipitation as hydroxides or sulfides at appropriate pH values, exploiting the very low Ksp values of these precipitates. Example: adjusting pH to 9-10 precipitates most heavy metals as hydroxides.

Frequently Asked Questions
1. How do you calculate molar solubility from Ksp? ⌄
Type MA (e.g., AgBr, AgCl, CaCO3): AB → A+ + B-; Ksp = s². s = sqrt(Ksp). Type MA2 or M2A (e.g., Zn(OH)2, CaF2, Ag2SO4): AB2 → A2+ + 2B-; Ksp = s(2s)² = 4s³. s = (Ksp/4)^(1/3). For A2B type: Ksp = (2s)²(s) = 4s³. Same formula! Type M2A2 (e.g., Hg2Cl2): M2X2 → M2^2+ + 2X-; Ksp = s(2s)² = 4s³. Same as MA2 type.
2. Why is Ksp only valid for sparingly soluble salts? ⌄
Ksp (solubility product) is an equilibrium constant for the dissolution of a sparingly soluble ionic solid. It is valid when: the concentration of dissolved ions is low enough that activities ≈ concentrations (ideal behaviour). For highly soluble salts (like NaCl, KNO3), the activity coefficients deviate significantly from 1 due to interionic interactions, and the simple Ksp relationship breaks down. Ksp is temperature-dependent (like all equilibrium constants) and is affected by common ions (common ion effect), pH (if one ion is acid/base active), and complex ion formation.
3. What is the common ion effect? ⌄
If a salt solution already contains one of the ions that the sparingly soluble salt would provide, the solubility of the salt decreases. Example: AgCl in 0.1 M NaCl (Na+ and Cl- ions). [Cl-] ≈ 0.1 M (from NaCl, >> from AgCl dissolution). Ksp(AgCl) = [Ag+][Cl-] = [Ag+] × 0.1 = 1.8×10^-10. [Ag+] = 1.8×10^-9 M. Compare to solubility in pure water: s = sqrt(1.8×10^-10) = 1.34×10^-5 M. In 0.1 M NaCl: s ≈ 1.8×10^-9 M — solubility reduced by factor ~7500! This is the common ion effect.
4. How does pH affect solubility? ⌄
For salts of weak acids (e.g., CaCO3, MgF2, Ca3(PO4)2): increasing acidity (lower pH) increases solubility because H+ consumes the anion (CO3^2- → HCO3- → H2CO3 → CO2; F- → HF). This is why limestone (CaCO3) dissolves in acid rain (H+ + CO3^2- → HCO3- → H2CO3). For metal hydroxides: increasing pH (more alkaline) decreases solubility (common ion OH- suppresses dissolution); decreasing pH increases solubility.
5. What is the relationship between Ksp and precipitate formation? ⌄
If the ionic product Q (= [M^n+][X^n-] at given concentrations) > Ksp: precipitate forms. Q < Ksp: no precipitate (solution is unsaturated, can dissolve more). Q = Ksp: saturated solution, equilibrium. Practical application: selective precipitation — by controlling [X-] added, you can precipitate one metal ion while leaving another in solution, separating them. Used in qualitative analysis (Group analysis with H2S, NaOH, etc.).
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