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Ksp and Solubility in Water Treatment Chemistry

By Aniket Bhardwaj · 4 September 2026 · Formula & Research

The solubility product is one of the first equilibrium constants a student meets, and it usually arrives attached to a dry question: "calculate the molar solubility of silver chloride." That same constant is the design tool behind municipal water softening, boiler scale control and industrial effluent treatment. This article works the numbers properly for two salts that matter in real plants, then explains why an engineer never trusts Ksp alone.

The formula

For AxBy(s) ⇌ x Ay+(aq) + y Bx−(aq)

Ksp = [Ay+]x [Bx−]y

and in terms of molar solubility s: Ksp = (x·s)x (y·s)y
TermMeaning
KspSolubility product — a constant at a given temperature; no unit is written
sMolar solubility: mol of the salt that dissolves per litre before saturation
QThe same product built from the actual concentrations in the water
Q < KspUnsaturated — more solid will dissolve
Q > KspSupersaturated — precipitation is thermodynamically favoured

The solid never appears in the expression. Its concentration is fixed by its own density, so it is folded into the constant.

Worked example 1 — how soluble is limestone?

Calcium carbonate, CaCO₃, has Ksp ≈ 3.3 × 10⁻⁹ at 25 °C. It dissolves 1 : 1, so [Ca²⁺] = [CO₃²⁻] = s.

Ksp = s × s = s²
s = √(3.3 × 10⁻⁹) = √(33 × 10⁻¹⁰) = 5.74 × 10⁻⁵ mol/L

Convert to a concentration a plant operator would recognise. The molar mass of CaCO₃ is 40.08 + 12.01 + (3 × 16.00) = 100.09 g/mol:
5.74 × 10⁻⁵ × 100.09 = 5.75 × 10⁻³ g/L = about 5.7 mg/L

That is very low — which is precisely why calcium carbonate is the material of choice for taking hardness out of water, and also why it is the commonest scale found inside pipes and boilers.

Worked example 2 — the common-ion effect

Now dissolve CaCO₃ not in pure water but in groundwater already containing 1.0 × 10⁻³ mol/L of calcium (roughly 40 mg/L Ca²⁺, an ordinary hard-water value).

The calcium already present dominates, so [Ca²⁺] ≈ 1.0 × 10⁻³ M.
[CO₃²⁻] = Ksp ÷ [Ca²⁺] = 3.3 × 10⁻⁹ ÷ 1.0 × 10⁻³ = 3.3 × 10⁻⁶ mol/L

Compare with 5.74 × 10⁻⁵ mol/L in pure water: solubility is suppressed by a factor of 5.74 × 10⁻⁵ ÷ 3.3 × 10⁻⁶ = about 17 times.

This is Le Chatelier's principle doing useful work. Adding one of the ions pushes the equilibrium back towards the solid, and a treatment plant exploits it deliberately: dosing extra carbonate or extra calcium drives more of the target ion out of solution than it could ever remove by simple settling.

Worked example 3 — a 1 : 2 salt

Magnesium hydroxide, Mg(OH)₂, has Ksp ≈ 5.6 × 10⁻¹². Here one formula unit gives one Mg²⁺ and two OH⁻, so the exponents change everything.

[Mg²⁺] = s and [OH⁻] = 2s, so
Ksp = s × (2s)² = 4s³
s³ = 5.6 × 10⁻¹² ÷ 4 = 1.4 × 10⁻¹²
s = ∛(1400 × 10⁻¹⁵) = 11.19 × 10⁻⁵ = 1.12 × 10⁻⁴ mol/L

Molar mass of Mg(OH)₂ = 24.31 + 2(16.00 + 1.008) = 58.32 g/mol, so s ≈ 1.12 × 10⁻⁴ × 58.32 = 6.5 mg/L.

At that saturation point [OH⁻] = 2s = 2.24 × 10⁻⁴ M, giving pOH = 3.65 and pH ≈ 10.35.

That pH figure is the whole reason magnesium removal is done the way it is: below about pH 10 there is simply not enough hydroxide for Mg(OH)₂ to be the stable phase, so lime must be dosed to push the pH up before magnesium hardness will come out.

Where this is actually used

Lime softening is the classic case. Slaked lime, Ca(OH)₂, is added to hard water and two reactions do the work:

Ca(HCO₃)₂ + Ca(OH)₂ → 2 CaCO₃↓ + 2 H₂O

Mg(HCO₃)₂ + 2 Ca(OH)₂ → Mg(OH)₂↓ + 2 CaCO₃↓ + 2 H₂O

Note the elegance of the first: calcium is added in order to remove calcium, because the added hydroxide converts bicarbonate into carbonate, and the carbonate then takes both the original and the added calcium out as insoluble solid. Ksp is what tells the designer how much lime is needed and what residual hardness is achievable.

The same reasoning appears wherever an unwanted ion must be turned into a solid:

The honest limitation: real water is not an ideal solution

Everything above assumes concentration equals chemical activity. In real water it does not, and the gap is not academic:

For those reasons practising engineers work with measured saturation indices, such as the Langelier Saturation Index, and with speciation software that carries activity corrections and ion pairing. Ksp is the concept underneath those tools — the first estimate and the sanity check, not the final answer.

Common mistakes

  • Forgetting the stoichiometric coefficient inside the bracket. For Mg(OH)₂ it is 4s³, not s³ or 2s³. Getting this wrong changed the answer by more than a factor of 1.5 in the example above.
  • Comparing Ksp values across different salt types. A 1 : 1 salt and a 1 : 2 salt cannot be ranked by Ksp alone, because the constants have different powers of concentration in them. Convert both to molar solubility first.
  • Using Q incorrectly. Q must be built from actual concentrations in the sample, with the same exponents as Ksp. Water with [Ca²⁺] = 2.0 × 10⁻³ and [CO₃²⁻] = 5.0 × 10⁻⁶ gives Q = 1.0 × 10⁻⁸, which exceeds 3.3 × 10⁻⁹ — scale is expected.
  • Ignoring the common ion. Solving s² = Ksp in water that already contains calcium overstates solubility roughly seventeen-fold, as shown above.
  • Assuming the tabulated Ksp applies at plant temperature. It is a 25 °C value unless stated otherwise.
  • Treating "insoluble" as absolute. Nothing has zero solubility. 5.7 mg/L is small, but over the volumes a treatment works handles it is a real mass.

Try your own salts. The Ksp calculator converts between solubility product and molar solubility for 1 : 1, 1 : 2 and 1 : 3 salts, so the 4s³ trap above is handled for you.

Open the Ksp / Solubility Calculator →

Ionic equilibria are a reliable scoring area in IIT-JAM, GATE, CSIR-NET and CUET-PG. ABC Chemistry runs coaching-centre and online chemical-sciences batches for students across India — abcchemistry.in.