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Chemistry · Equilibrium

Solubility Product & Common Ion Effect

When a sparingly soluble salt sits in contact with its saturated solution, a true dynamic equilibrium is established between the undissolved solid and its ions. The equilibrium constant of that process — the solubility product $K_{sp}$ — together with the common ion effect, governs how much salt dissolves and when a precipitate appears. Drawn from NCERT Class 11 Chemistry, Unit 6 (sections 6.13.1–6.13.2) and supplemented by the NIOS Ionic Equilibrium chapter, this is one of the most reliably examined corners of NEET equilibrium.

The Saturated-Solution Equilibrium

Every "insoluble" salt is in truth only sparingly soluble: a small but definite quantity passes into solution as ions. Once the solution becomes saturated, the rate at which ions leave the crystal lattice equals the rate at which they return to it, and a heterogeneous equilibrium is set up. Taking barium sulphate as NCERT's reference case, the equilibrium between the undissolved solid and the ions in its saturated solution is written as

$$\ce{BaSO4(s) <=> Ba^2+(aq) + SO4^2-(aq)}$$

This is a genuine equilibrium, not a one-way process. The schematic below captures the dynamic balance: solid is continually dissolving while dissolved ions are continually re-depositing, with no net change in the amount of solid at saturation.

Figure 1

Dynamic equilibrium at the surface of a sparingly soluble salt.

Saturated solution BaSO₄(s) Ba²⁺ SO₄²⁻ dissolve deposit

Solubility Product Constant Ksp

For the barium sulphate equilibrium, the equilibrium constant would formally be written with the solid in the denominator:

$$K = \frac{[\ce{Ba^2+}][\ce{SO4^2-}]}{[\ce{BaSO4}]}$$

The concentration (effective activity) of a pure solid is a constant, so it is folded into the constant itself. What survives is the product of the ionic concentrations alone:

$$K_{sp} = K\,[\ce{BaSO4}] = [\ce{Ba^2+}][\ce{SO4^2-}]$$

This $K_{sp}$ is the solubility product constant. For $\ce{BaSO4}$ at 298 K its experimental value is $1.1 \times 10^{-10}$. The defining feature is that each ionic concentration is raised to the power of its stoichiometric coefficient — a fact that becomes decisive the moment a salt produces more than one ion of either species.

Ksp and Molar Solubility S

Molar solubility $S$ is the number of moles of the salt that dissolve per litre to give a saturated solution. The link between $K_{sp}$ and $S$ follows directly from the dissolution stoichiometry. For a 1:1 (AB-type) salt such as $\ce{BaSO4}$, each formula unit yields one cation and one anion, so $[\ce{Ba^2+}] = [\ce{SO4^2-}] = S$ and

$$K_{sp} = (S)(S) = S^2 \quad\Rightarrow\quad S = \sqrt{K_{sp}}$$

For barium sulphate this gives $S = \sqrt{1.1\times10^{-10}} = 1.05\times10^{-5}\ \text{mol L}^{-1}$. The picture changes for salts of higher stoichiometry. NCERT generalises to a salt $\ce{M_{x}X_{y}}$:

$$\ce{M_{x}X_{y}(s) <=> $x$ M^{p+}(aq) + $y$ X^{q-}(aq)}$$

Here $[\ce{M^{p+}}] = xS$ and $[\ce{X^{q-}}] = yS$, so the general relation is

$$K_{sp} = (xS)^x (yS)^y = x^{x}\,y^{y}\,S^{(x+y)} \quad\Rightarrow\quad S = \left(\frac{K_{sp}}{x^{x}\,y^{y}}\right)^{1/(x+y)}$$
NEET Trap

Equal Ksp does not mean equal solubility

Because $S$ enters $K_{sp}$ with a different power for each salt type, two salts with identical $K_{sp}$ can have very different solubilities. For an AB salt $S=\sqrt{K_{sp}}$; for an AB₂ salt $S=(K_{sp}/4)^{1/3}$. Never rank solubilities from raw $K_{sp}$ values unless the salts are of the same formula type. Likewise, the common ion effect always lowers molar solubility — never raises it.

Compare $S$, not $K_{sp}$, when the salt formulas differ.

Master Table: Salt Type to Ksp

The following table consolidates the $K_{sp}$–$S$ relationships for the salt types NEET examines most often. Memorising the right-hand column converts almost every numerical solubility question into a one-line substitution.

Salt Type Dissolution Equilibrium Ksp in terms of S Solubility S Example
AB $\ce{AB(s) <=> A+ + B-}$ $K_{sp} = S^2$ $S = K_{sp}^{1/2}$ $\ce{AgCl},\ \ce{BaSO4}$
AB₂ $\ce{AB2(s) <=> A^2+ + 2B-}$ $K_{sp} = 4S^3$ $S = (K_{sp}/4)^{1/3}$ $\ce{CaF2},\ \ce{Mg(OH)2}$
A₂B $\ce{A2B(s) <=> 2A+ + B^2-}$ $K_{sp} = 4S^3$ $S = (K_{sp}/4)^{1/3}$ $\ce{Ag2CrO4},\ \ce{Ag2C2O4}$
AB₃ $\ce{AB3(s) <=> A^3+ + 3B-}$ $K_{sp} = 27S^4$ $S = (K_{sp}/27)^{1/4}$ $\ce{Fe(OH)3},\ \ce{AlCl3}$ (type)
A₂B₃ $\ce{A2B3(s) <=> 2A^3+ + 3B^2-}$ $K_{sp} = 108S^5$ $S = (K_{sp}/108)^{1/5}$ $\ce{A2X3}$ (NCERT Problem 6.26)
MₓXₙ (general) $\ce{M_{x}X_{y}(s) <=> $x$M^{p+} + $y$X^{q-}}$ $K_{sp} = x^{x}y^{y}S^{(x+y)}$ $S = \left(\dfrac{K_{sp}}{x^{x}y^{y}}\right)^{\!1/(x+y)}$ Zirconium phosphate, $6912\,S^7$

Worked Examples

Two solved problems, both adapted from the NCERT in-text exercises, show how the master table is deployed. Note how the cube and the higher-root structures appear naturally from the stoichiometry.

Example 1 · A₂X₃ type

Calculate the solubility of $\ce{A2X3}$ in pure water, given $K_{sp} = 1.1 \times 10^{-23}$, assuming neither ion reacts with water.

The salt dissociates as $\ce{A2X3 -> 2A^3+ + 3X^2-}$. If $S$ is the solubility, then $[\ce{A^3+}] = 2S$ and $[\ce{X^2-}] = 3S$.

$$K_{sp} = (2S)^2 (3S)^3 = 4S^2 \cdot 27S^3 = 108\,S^5 = 1.1 \times 10^{-23}$$

Thus $S^5 = 1.0 \times 10^{-25}$, giving $\;S = 1.0 \times 10^{-5}\ \text{mol L}^{-1}$. This matches the $108S^5$ row of the master table exactly.

Example 2 · Comparing two salts

$K_{sp}$ of $\ce{Ni(OH)2}$ is $2.0 \times 10^{-15}$ and of $\ce{AgCN}$ is $6 \times 10^{-17}$. Which salt is more soluble?

$\ce{AgCN <=> Ag+ + CN-}$ is an AB salt: $K_{sp} = S_1^2 = 6\times10^{-17}$, so $S_1 = 7.8\times10^{-9}\ \text{mol L}^{-1}$.

$\ce{Ni(OH)2 <=> Ni^2+ + 2OH-}$ is an AB₂ salt: $K_{sp} = (S_2)(2S_2)^2 = 4S_2^3 = 2.0\times10^{-15}$, so $S_2 = 0.58\times10^{-4}\ \text{mol L}^{-1}$.

Despite its larger $K_{sp}$, $\ce{Ni(OH)2}$ is the more soluble salt — a direct illustration of the NEET trap above.

The Common Ion Effect

The common ion effect is defined in NCERT as a shift in equilibrium on adding a substance that provides more of an ionic species already present in the dissociation equilibrium. Applied to a saturated salt solution, Le Chatelier's principle predicts the consequence at once: increasing the concentration of one ion drives it to combine with the oppositely charged ion, and salt precipitates until $K_{sp} = Q_{sp}$ is restored.

Because $K_{sp}$ is fixed at a given temperature, forcing up one ion's concentration must force down the other ion's concentration — and with it the molar solubility. The figure contrasts the dissolution of $\ce{AgCl}$ in pure water with its dissolution in a chloride-rich medium.

Figure 2

Common-ion suppression of solubility for $\ce{AgCl}$.

AgCl in pure water solubility = S AgCl in NaCl (Cl⁻ added) solubility < S = Cl⁻ = Ag⁺

The same logic applies even to highly soluble salts. Passing $\ce{HCl}$ gas through a saturated solution of $\ce{NaCl}$ raises the chloride concentration so sharply that $\ce{NaCl}$ precipitates out — the standard laboratory route to ultrapure sodium chloride, freed of impurities such as sodium and magnesium sulphates.

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The common ion effect also suppresses the ionisation of weak acids and bases — the foundation of buffers. See pH, Ionisation & Buffer Solutions.

Ionic Product & Predicting Precipitation

The expression $[\ce{M^{p+}}]^x[\ce{X^{q-}}]^y$ evaluated for any set of concentrations — not necessarily equilibrium ones — is called the ionic product, $Q_{sp}$. It has the same form as $K_{sp}$ but is a snapshot of the current state rather than the saturation value. The comparison of $Q_{sp}$ with $K_{sp}$ tells us the direction of change.

ConditionState of SolutionWhat Happens
$Q_{sp} < K_{sp}$UnsaturatedMore solid can dissolve; no precipitate.
$Q_{sp} = K_{sp}$Saturated (equilibrium)Solution exactly saturated; no net change.
$Q_{sp} > K_{sp}$SupersaturatedSalt precipitates until $Q_{sp}$ falls to $K_{sp}$.

When two solutions are mixed, dilution must be accounted for before $Q_{sp}$ is computed: each ion's concentration is its moles divided by the total combined volume. Only then is $Q_{sp}$ compared with $K_{sp}$ to decide whether a precipitate appears.

NEET Trap

Forgetting dilution before comparing Qsp with Ksp

In a "will a precipitate form?" question, the most common error is plugging in the original concentrations of the two solutions. Mixing equal volumes halves each ion's concentration. Compute the post-mixing concentrations first, then form $Q_{sp}$, then compare with $K_{sp}$. A precipitate forms only when $Q_{sp} > K_{sp}$.

$Q_{sp} > K_{sp} \Rightarrow$ precipitation; $Q_{sp} \le K_{sp} \Rightarrow$ no precipitate.

Applications in Qualitative Analysis

The interplay of $K_{sp}$ and the common ion effect underpins both gravimetric estimation and the systematic qualitative analysis of cations. By deliberately supplying a common ion of very low-$K_{sp}$ partner, an ion can be precipitated almost completely for accurate weighing.

Ion to be precipitatedPrecipitated asEquilibrium
$\ce{Ag+}$Silver chloride$\ce{Ag+ + Cl- <=> AgCl(v)}$
$\ce{Fe^3+}$Ferric hydroxide (hydrated oxide)$\ce{Fe^3+ + 3OH- <=> Fe(OH)3(v)}$
$\ce{Ba^2+}$Barium sulphate$\ce{Ba^2+ + SO4^2- <=> BaSO4(v)}$

In the classical scheme, controlling the concentration of a precipitating ion through a common ion (for example, suppressing sulphide ion by adding acid) allows the selective separation of metal-ion groups whose sulphides differ in $K_{sp}$. The salt-analysis groundwork sits in a separate chapter; here the takeaway is conceptual — precipitation is engineered, not accidental, once $K_{sp}$ and $Q_{sp}$ are understood.

Quick Recap

Five lines to carry into the exam hall

  • $K_{sp} = [\ce{M^{p+}}]^x[\ce{X^{q-}}]^y$ at saturation; the pure solid is excluded from the expression.
  • AB: $K_{sp}=S^2$. AB₂ / A₂B: $K_{sp}=4S^3$. AB₃: $27S^4$. A₂B₃: $108S^5$. General: $x^{x}y^{y}S^{(x+y)}$.
  • Equal $K_{sp}$ ≠ equal solubility — compare $S$ when formula types differ.
  • Common ion effect lowers the solubility of a sparingly soluble salt (Le Chatelier).
  • $Q_{sp} > K_{sp}$ precipitate; $Q_{sp} = K_{sp}$ saturated; $Q_{sp} < K_{sp}$ unsaturated. Dilute first when mixing.

NEET PYQ Snapshot — Solubility Product & Common Ion Effect

Real NEET questions on the Ksp–solubility relation, common ion suppression and precipitation logic.

NEET 2018

The solubility of $\ce{BaSO4}$ in water is $2.42 \times 10^{-3}\ \text{g L}^{-1}$ at 298 K. The value of its solubility product $K_{sp}$ will be (molar mass of $\ce{BaSO4} = 233\ \text{g mol}^{-1}$):

  1. $1.08 \times 10^{-10}\ \text{mol}^2\text{L}^{-2}$
  2. $1.08 \times 10^{-12}\ \text{mol}^2\text{L}^{-2}$
  3. $1.08 \times 10^{-14}\ \text{mol}^2\text{L}^{-2}$
  4. $1.08 \times 10^{-8}\ \text{mol}^2\text{L}^{-2}$
Answer: (1)

Convert mass solubility to molar: $S = \dfrac{2.42\times10^{-3}}{233} = 1.03\times10^{-5}\ \text{mol L}^{-1}$. For the AB salt $\ce{BaSO4}$, $K_{sp} = S^2 = (1.03\times10^{-5})^2 = 1.08\times10^{-10}$.

NEET 2017

Concentration of the $\ce{Ag+}$ ions in a saturated solution of $\ce{Ag2C2O4}$ is $2.2 \times 10^{-4}\ \text{mol L}^{-1}$. Solubility product of $\ce{Ag2C2O4}$ is:

  1. $5.3 \times 10^{-12}$
  2. $2.42 \times 10^{-8}$
  3. $2.66 \times 10^{-12}$
  4. $4.5 \times 10^{-11}$
Answer: (1)

$\ce{Ag2C2O4 <=> 2Ag+ + C2O4^2-}$ (A₂B type). Here $[\ce{Ag+}] = 2S = 2.2\times10^{-4}$, so $S = 1.1\times10^{-4}$. Then $K_{sp} = 4S^3 = 4 \times (1.1\times10^{-4})^3 = 5.3\times10^{-12}$.

NEET 2016

MY and NY₃, two nearly insoluble salts, have the same $K_{sp}$ values of $6.2 \times 10^{-13}$ at room temperature. Which statement would be true in regard to MY and NY₃?

  1. The molar solubility of MY in water is less than that of NY₃
  2. The salts MY and NY₃ are more soluble in 0.5 M KY than in pure water
  3. The addition of the salt KY to solutions of MY and NY₃ will have no effect on their solubilities
  4. The molar solubilities of MY and NY₃ in water are identical
Answer: (1)

MY (AB): $K_{sp}=S^2 \Rightarrow S = \sqrt{6.2\times10^{-13}} = 8\times10^{-7}$. NY₃ (AB₃): $K_{sp}=27S^4 \Rightarrow S = (6.2\times10^{-13}/27)^{1/4} = 3.89\times10^{-4}$. So $S_{MY} < S_{NY_3}$ — equal $K_{sp}$, unequal solubility. Option (2) is wrong because the common ion KY lowers solubility.

NEET 2020

Find out the solubility of $\ce{Ni(OH)2}$ in 0.1 M NaOH. Given that the ionic product of $\ce{Ni(OH)2}$ is $2 \times 10^{-15}$:

  1. $2 \times 10^{-8}\ \text{M}$
  2. $1 \times 10^{-13}\ \text{M}$
  3. $1 \times 10^{8}\ \text{M}$
  4. $2 \times 10^{-13}\ \text{M}$
Answer: (4)

NaOH supplies $0.1\ \text{M}\ \ce{OH-}$ (the common ion). For $\ce{Ni(OH)2 <=> Ni^2+ + 2OH-}$, with solubility $S$: total $[\ce{OH-}] = (0.1 + 2S) \approx 0.1$. So $K_{sp} = S(0.1)^2 = 2\times10^{-15}$, giving $S = 2\times10^{-13}\ \text{M}$ — drastically below its value in pure water, the common ion effect in action.

FAQs — Solubility Product & Common Ion Effect

The recurring conceptual snags from this subtopic, answered tightly.

What is the difference between solubility product (Ksp) and ionic product (Qsp)?

Both are products of ionic concentrations raised to their stoichiometric powers, but Ksp is the value only at saturation (equilibrium), whereas Qsp (the ionic product) is computed for any arbitrary set of concentrations. At equilibrium Ksp = Qsp. When Qsp < Ksp the solution is unsaturated and more solid can dissolve; when Qsp > Ksp the solution is supersaturated and the salt precipitates until Qsp falls back to Ksp.

How is Ksp related to molar solubility S for different salt types?

For an AB salt such as AgCl, Ksp = S². For an AB2 (or A2B) salt such as CaF2 or Ag2CrO4, Ksp = 4S³. For a general salt MxXy, Ksp = xˣ·yʸ·S^(x+y). So you cannot compare solubilities directly from Ksp values unless the salts share the same formula type — different powers of S are involved.

Why does the common ion effect lower the solubility of a sparingly soluble salt?

By Le Chatelier's principle, adding an ion already present in the dissolution equilibrium raises Qsp above Ksp, so the system shifts toward the undissolved solid and the salt precipitates until Qsp = Ksp again. Because Ksp is fixed at a given temperature, increasing one ion's concentration forces the other ion's concentration — and hence the molar solubility — to fall.

Two salts MY and NY3 have the same Ksp. Which is more soluble?

They are not equally soluble despite equal Ksp, because the Ksp–S relation differs. For MY (AB type), Ksp = S², so S = √Ksp. For NY3 (AB3 type), Ksp = 27S⁴, so S = (Ksp/27)^(1/4). For a small Ksp value such as 6.2 × 10⁻¹³, the AB3 salt turns out to be the more soluble of the two; the molar solubility of MY is less than that of NY3.

How do you predict whether a precipitate will form when two solutions are mixed?

First account for dilution to get the concentrations of the relevant ions after mixing. Compute the ionic product Qsp using those mixed concentrations. Compare with Ksp: if Qsp > Ksp a precipitate forms; if Qsp = Ksp the solution is exactly saturated; if Qsp < Ksp no precipitate forms and the solution remains unsaturated.

Why is the concentration of the pure solid not included in the Ksp expression?

The concentration (effective activity) of a pure solid is constant and does not change as the salt dissolves. It is therefore absorbed into the equilibrium constant: Ksp = K × [solid]. As a result, Ksp depends only on the product of the ionic concentrations in the saturated solution, each raised to its stoichiometric coefficient.

Related Topics
Equilibrium Return to the full chapter hub for all equilibrium subtopics. pH, Ionisation & Buffers Common ion effect applied to weak acids, bases and buffer action. Hydrolysis of Salts & pH How dissolved salts shift the pH of their solutions. Ionic Equilibrium: Acids & Bases Ka, Kb, degree of ionisation and the broader ionic equilibrium picture. Salt Analysis Qualitative cation/anion separation where Ksp control is applied.