Solubility Equilibria of Sparingly Soluble Salts — Revision Notes

⚡ 30-Second Revision

Equilibrium: — $A_x B_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)$
$K_{sp}$ expression: — $K_{sp} = [A^{y+}]^x [B^{x-}]^y$
$s$ vs $K_{sp}$ relations:**

- AB type: $K_{sp} = s^2 \implies s = \sqrt{K_{sp}}$ - $AB_2$ or $A_2B$ type: $K_{sp} = 4s^3 \implies s = \sqrt[3]{K_{sp}/4}$ - $A_x B_y$ type: $K_{sp} = x^x y^y s^{(x+y)}$

Precipitation condition:

- $Q_{sp} < K_{sp}$: Unsaturated, no precipitation - $Q_{sp} = K_{sp}$: Saturated, equilibrium - $Q_{sp} > K_{sp}$: Supersaturated, precipitation occurs

Common Ion Effect: — Decreases solubility ($s$), $K_{sp}$ remains constant.
pH Effect: — Increases solubility for salts with basic anions in acidic solutions (e.g., $Mg(OH)_2$, $CaCO_3$, $PbS$).
Complex Ion Effect: — Increases solubility if one ion forms a stable complex (e.g., AgCl in $NH_3$).

2-Minute Revision

Solubility equilibria describe the dynamic balance between a sparingly soluble ionic solid and its ions in a saturated solution. The key quantitative measure is the solubility product constant, $K_{sp}$, which is an equilibrium constant specific to a given salt at a particular temperature.

For a salt $A_x B_y$, $K_{sp} = [A^{y+}]^x [B^{x-}]^y$. Molar solubility ($s$) is the concentration of the dissolved salt and is related to $K_{sp}$ by stoichiometry (e.g., $s = \sqrt{K_{sp}}$ for AB salts, $s = \sqrt[3]{K_{sp}/4}$ for $AB_2$ salts).

Precipitation is predicted by comparing the ion product ($Q_{sp}$) with $K_{sp}$: if $Q_{sp} > K_{sp}$, precipitation occurs. Factors influencing solubility include the common ion effect (decreases solubility), pH (increases solubility for salts with basic anions in acidic conditions), and complex ion formation (increases solubility).

Remember, $K_{sp}$ itself only changes with temperature, while solubility ($s$) can change due to solution composition.

5-Minute Revision

Solubility equilibria are fundamental to understanding how sparingly soluble ionic compounds behave in solution. A sparingly soluble salt, despite its name, does dissolve to a small extent, establishing a dynamic equilibrium between the undissolved solid and its constituent ions.

This equilibrium is quantified by the solubility product constant, $K_{sp}$. For a generic salt $A_x B_y$, the dissolution equilibrium is $A_x B_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)$, and its $K_{sp}$ is expressed as $K_{sp} = [A^{y+}]^x [B^{x-}]^y$.

The molar solubility ($s$) of the salt is the concentration of the dissolved species in a saturated solution. The relationship between $s$ and $K_{sp}$ depends on the salt's stoichiometry: for AB type salts, $K_{sp} = s^2$; for $AB_2$ or $A_2B$ type salts, $K_{sp} = 4s^3$; and for $A_x B_y$ type, $K_{sp} = x^x y^y s^{(x+y)}$.

Predicting precipitation is a crucial application. We calculate the ion product ($Q_{sp}$) using the current (non-equilibrium) concentrations of ions. If $Q_{sp} < K_{sp}$, the solution is unsaturated; if $Q_{sp} = K_{sp}$, it's saturated; and if $Q_{sp} > K_{sp}$, precipitation will occur until equilibrium is re-established. Several factors can influence the solubility of a sparingly soluble salt:

1
Common Ion Effect: — Adding a soluble salt containing an ion common to the sparingly soluble salt (e.g., adding NaCl to AgCl solution) shifts the equilibrium towards the solid, decreasing the solubility of the sparingly soluble salt. This is a direct application of Le Chatelier's Principle.
2
Effect of pH: — The solubility of salts with basic anions (e.g., $CO_3^{2-}$, $S^{2-}$, $OH^-$) increases in acidic solutions because $H^+$ ions react with the basic anion, removing it from solution and shifting the dissolution equilibrium to the right. For example, $Mg(OH)_2$ is more soluble in acid.
3
Complex Ion Formation: — If one of the ions of the sparingly soluble salt can form a stable complex ion with a ligand present in the solution, its concentration in the free state decreases. This shifts the dissolution equilibrium to the right, increasing the solubility of the salt. For instance, AgCl dissolves in ammonia due to the formation of $[Ag(NH_3)_2]^+$.

Remember that $K_{sp}$ is a constant at a given temperature, while solubility ($s$) is a variable that changes with solution composition. Mastering these concepts and their quantitative relationships is vital for NEET.

Prelims Revision Notes

Solubility Equilibria: Key Facts for NEET

1
Sparingly Soluble Salts: — These are ionic compounds that dissolve to a very small extent in water, forming a saturated solution in dynamic equilibrium with the undissolved solid.

* Example: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$

1
**Solubility Product Constant ($K_{sp}$):**

* It's an equilibrium constant for the dissolution of a sparingly soluble salt. * For $A_x B_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)$, $K_{sp} = [A^{y+}]^x [B^{x-}]^y$. * The concentration of the pure solid is omitted from the expression. * $K_{sp}$ is temperature-dependent; it does NOT change with common ion or pH.

1
**Molar Solubility ($s$):**

* Expressed in mol/L, it's the concentration of the dissolved salt in a saturated solution. * **Relationship between $s$ and $K_{sp}$:** * AB type (e.g., AgCl): $K_{sp} = s^2 \implies s = \sqrt{K_{sp}}$ * **$AB_2$ or $A_2B$ type** (e.g., $CaF_2$, $Ag_2CrO_4$): $K_{sp} = 4s^3 \implies s = \sqrt[3]{K_{sp}/4}$ * **$A_3B_2$ type** (e.g., $Ca_3(PO_4)_2$): $K_{sp} = (3s)^3(2s)^2 = 108s^5$ * **General $A_x B_y$ type:** $K_{sp} = x^x y^y s^{(x+y)}$

1
**Ion Product ($Q_{sp}$):**

* Calculated like $K_{sp}$ but using *initial* (non-equilibrium) ion concentrations. * Predicting Precipitation: * $Q_{sp} < K_{sp}$: Unsaturated, no precipitation. * $Q_{sp} = K_{sp}$: Saturated, at equilibrium. * $Q_{sp} > K_{sp}$: Supersaturated, precipitation occurs.

1
Factors Affecting Solubility (Le Chatelier's Principle):

* Common Ion Effect: Adding a common ion (e.g., $Cl^-$ to AgCl) decreases the solubility of the sparingly soluble salt by shifting the equilibrium to the left (towards solid formation). * Effect of pH: * Salts with basic anions (conjugate bases of weak acids, e.

g., $CO_3^{2-}$, $S^{2-}$, $OH^-$, $F^-$) become *more soluble* in acidic solutions because $H^+$ reacts with the anion, removing it from solution. * Salts with anions of strong acids (e.g., $Cl^-$, $Br^-$, $I^-$, $NO_3^-$, $SO_4^{2-}$) are generally *unaffected* by pH.

* Complex Ion Formation: If a metal ion can form a stable complex with a ligand (e.g., $Ag^+$ with $NH_3$), its concentration of free ions decreases, shifting the equilibrium to the right and *increasing* solubility.

* Temperature: Most dissolution processes are endothermic, so increasing temperature generally increases solubility. $K_{sp}$ values are temperature-dependent.

Key Calculation Steps:

1
Write balanced dissolution equation.
2
Write $K_{sp}$ expression.
3
Relate $s$ to ion concentrations based on stoichiometry.
4
Substitute into $K_{sp}$ and solve (or vice-versa).
5
For common ion effect, use initial common ion concentration and approximate if $s$ is small.
6
For $Q_{sp}$ calculations, remember dilution effects when mixing solutions.

Vyyuha Quick Recall

SPARINGLY SOLUBLE SALTS: Shift PH, Add Reagents, Ion Numbers, Get Lower Yields.

Shift PH: Solubility changes with pH for salts with basic anions.
Add Reagents: Common ion effect (adding a common ion) decreases solubility.
Ion Numbers: Stoichiometry is crucial for $K_{sp}$ and $s$ relationship (e.g., $s^2$, $4s^3$).
Get Lower Yields: $Q_{sp} > K_{sp}$ means precipitation, reducing ion yield in solution.