Chemistry

Solubility Equilibria of Sparingly Soluble Salts

Chemistry·Revision Notes

Common Ion Effect — Revision Notes

NEET UG
Version 1Updated 22 Mar 2026

⚡ 30-Second Revision

  • Definition:Decrease in solubility of a sparingly soluble salt by adding a soluble salt with a common ion.
  • Principle:Le Chatelier's Principle.
  • Equilibrium:MaXb(s)aMb+(aq)+bXa(aq)M_aX_b(s) \rightleftharpoons aM^{b+}(aq) + bX^{a-}(aq)
  • $K_{sp}$:Remains constant at a given temperature.
  • Effect:Shifts equilibrium left, reducing solubility.
  • Calculation:For MX(s)MX(s), Ksp=[M+][X]K_{sp} = [M^+][X^-]. If common ion [X]C[X^-]_C is added, then [X]total[X]C[X^-]_{\text{total}} \approx [X^-]_C. Solubility s=[M+]=Ksp/[X]Cs = [M^+] = K_{sp}/[X^-]_C.
  • Approximation:s[CommonIon]initials \ll [Common\,Ion]_{\text{initial}} is often valid.
  • pH Effect:For salts with basic anions (e.g., OHOH^-, CO32CO_3^{2-}), adding acid increases solubility (removes anion). Adding base (common OHOH^-) decreases solubility of hydroxides.

2-Minute Revision

The Common Ion Effect is a crucial concept in ionic equilibrium, explaining why a sparingly soluble ionic compound becomes even less soluble when a soluble salt containing one of its constituent ions (a 'common ion') is added to the solution.

This phenomenon is a direct consequence of Le Chatelier's Principle. When the concentration of a product ion in a solubility equilibrium is increased by adding a common ion, the equilibrium shifts to the left, favoring the formation of the undissolved solid and thus reducing the molar solubility of the sparingly soluble salt.

It's vital to remember that the solubility product constant (KspK_{sp}) itself does not change; only the equilibrium concentrations of the ions adjust to maintain the constant KspK_{sp} value. Quantitative problems typically involve calculating the new solubility by using the KspK_{sp} expression and making the approximation that the contribution of the sparingly soluble salt to the common ion concentration is negligible.

This effect is widely applied in selective precipitation and purification processes, and understanding its interplay with pH is also important for NEET.

5-Minute Revision

Let's consolidate the Common Ion Effect, a cornerstone of ionic equilibrium for NEET. It describes the reduction in solubility of a sparingly soluble salt when a soluble salt providing a common ion is introduced.

For example, consider AgCl(s)Ag+(aq)+Cl(aq)AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq). If we add NaClNaCl, which provides ClCl^- ions, the increased [Cl][Cl^-] shifts the equilibrium to the left, causing more AgClAgCl to precipitate and reducing its solubility.

This is a direct application of Le Chatelier's Principle.

Key Points & Calculations:

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  1. $K_{sp}$ is constant:The solubility product constant (KspK_{sp}) for the sparingly soluble salt remains unchanged at a given temperature. The system simply adjusts ion concentrations to maintain KspK_{sp}.
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  3. Molar Solubility (s):In pure water, for a 1:11:1 salt MXMX, s=Ksps = \sqrt{K_{sp}}. For MX2MX_2, s=(Ksp/4)1/3s = (K_{sp}/4)^{1/3}.
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  5. In presence of common ion:Let's say we have MX(s)MX(s) and add a common ion XX^- from a soluble salt at concentration CC. The equilibrium concentrations become [M+]=s[M^+] = s' and [X]=s+C[X^-] = s' + C. Since ss' is usually very small, we approximate [X]C[X^-] \approx C. Then Ksp=(s)(C)K_{sp} = (s')(C), so s=Ksp/Cs' = K_{sp}/C. This ss' is the new, reduced solubility.

Example: Ksp(AgCl)=1.8×1010K_{sp}(AgCl) = 1.8 \times 10^{-10}. In 0.1M0.1\,M NaClNaCl, s=1.8×10100.1=1.8×109Ms' = \frac{1.8 \times 10^{-10}}{0.1} = 1.8 \times 10^{-9}\,M. (Compare to s=1.34×105Ms = 1.34 \times 10^{-5}\,M in pure water).

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  1. pH Effects:For salts like Mg(OH)2(s)Mg2+(aq)+2OH(aq)Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq), if pH is controlled (e.g., by a buffer), the [OH][OH^-] is fixed. This fixed [OH][OH^-] acts as a common ion, reducing Mg(OH)2Mg(OH)_2 solubility. If acid is added, it reacts with OHOH^-, removing it and *increasing* solubility (not a common ion effect, but a related equilibrium shift).
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  3. Selective Precipitation:To separate ions (e.g., Ca2+Ca^{2+} and Ba2+Ba^{2+}), calculate the minimum common ion concentration needed to precipitate each. The ion requiring the lowest concentration precipitates first.

Always identify the common ion, set up the KspK_{sp} expression correctly, and apply the approximation where appropriate to save time. This topic is frequently tested in NEET.

Prelims Revision Notes

Common Ion Effect: NEET Revision Notes

1. Definition & Principle:

  • Common Ion Effect:Decrease in the solubility of a sparingly soluble ionic compound when a soluble salt containing a common ion is added to the solution.
  • Underlying Principle:Le Chatelier's Principle. Adding a product ion shifts the equilibrium towards the reactants (undissolved solid).

2. Solubility Equilibrium & $K_{sp}$:

  • For a sparingly soluble salt MaXb(s)aMb+(aq)+bXa(aq)M_aX_b(s) \rightleftharpoons aM^{b+}(aq) + bX^{a-}(aq).
  • Solubility Product Constant ($K_{sp}$):Ksp=[Mb+]a[Xa]bK_{sp} = [M^{b+}]^a[X^{a-}]^b.
  • Crucial:KspK_{sp} is constant at a given temperature; it does NOT change due to the common ion effect. Only the equilibrium concentrations of ions change.

3. Quantitative Aspects (Calculations):

  • Molar Solubility (s):Moles of salt dissolved per liter of solution.
  • In Pure Water:

* For MX(s)M+(aq)+X(aq)MX(s) \rightleftharpoons M^+(aq) + X^-(aq), Ksp=s2    s=KspK_{sp} = s^2 \implies s = \sqrt{K_{sp}}. * For MX2(s)M2+(aq)+2X(aq)MX_2(s) \rightleftharpoons M^{2+}(aq) + 2X^-(aq), Ksp=(s)(2s)2=4s3    s=(Ksp/4)1/3K_{sp} = (s)(2s)^2 = 4s^3 \implies s = (K_{sp}/4)^{1/3}. * For M2X(s)2M+(aq)+X2(aq)M_2X(s) \rightleftharpoons 2M^+(aq) + X^{2-}(aq), Ksp=(2s)2(s)=4s3    s=(Ksp/4)1/3K_{sp} = (2s)^2(s) = 4s^3 \implies s = (K_{sp}/4)^{1/3}.

  • In Presence of Common Ion:

1. Identify the common ion and its concentration (CC) from the added soluble salt. 2. Let ss' be the new molar solubility of the sparingly soluble salt. 3. Set up the KspK_{sp} expression, accounting for the common ion concentration.

* Example (MXMX in CXCX solution): [M+]=s[M^+] = s', [X]total=s+C[X^-]_{\text{total}} = s' + C. * Ksp=(s)(s+C)K_{sp} = (s')(s' + C). 4. Approximation: Since ss' is usually very small compared to CC, assume s+CCs' + C \approx C.

* Then Ksp=(s)(C)    s=Ksp/CK_{sp} = (s')(C) \implies s' = K_{sp}/C. 5. Check Approximation: After calculating ss', verify if sCs' \ll C (e.g., ss' is less than 5% of CC). If not, solve the quadratic equation (rarely needed for NEET).

4. Applications & Related Concepts:

  • Selective Precipitation:Used to separate ions from a mixture by adding a common precipitating agent. The ion with the lower required common ion concentration precipitates first.
  • Purification:Common ion effect is used to precipitate pure salts from impure solutions (e.g., NaClNaCl purification using HClHCl).
  • pH Effect on Solubility:For salts whose ions are weak acids or bases (e.g., Mg(OH)2Mg(OH)_2, CaCO3CaCO_3, ZnSZnS), changing pH can significantly alter solubility.

* If anion is basic (OHOH^-, CO32CO_3^{2-}): Adding acid (H+H^+) removes the anion, increasing solubility. Adding base (common OHOH^-) decreases solubility. * If cation is acidic: Adding base removes H+H^+, increasing solubility.

5. Common Pitfalls:

  • Confusing KspK_{sp} with solubility. KspK_{sp} is constant, solubility changes.
  • Forgetting stoichiometry in KspK_{sp} expression (e.g., (2s)2(2s)^2 for MX2MX_2).
  • Not applying the common ion approximation correctly or when it's valid.
  • Confusing Common Ion Effect (decreases solubility) with Salt Effect (increases solubility due to inert ions).

Vyyuha Quick Recall

Common Ion Effect: Causes Ions to Exit (precipitate)!

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