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Equilibrium - Solubility Product Constant

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Solubility Product Constant (KspK_{sp}) is a type of equilibrium constant that applies to sparingly soluble salts in a saturated solution.

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For a sparingly soluble salt AxByA_x B_y in equilibrium with its saturated solution: AxBy(s)ightleftharpoonsxAy+(aq)+yBx−(aq)A_x B_y (s) ightleftharpoons x A^{y+} (aq) + y B^{x-} (aq).

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Molar solubility (ss) is defined as the number of moles of the solute dissolved per litre of its saturated solution at a particular temperature.

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The Common Ion Effect: The solubility of a sparingly soluble salt decreases if a soluble salt containing a common ion is added to the solution, shifting the equilibrium towards the undissociated solid.

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The Ionic Product (QspQ_{sp}) is calculated using the same expression as KspK_{sp}, but using initial or instantaneous concentrations instead of equilibrium concentrations.

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Criteria for precipitation: If Qsp>KspQ_{sp} > K_{sp}, the solution is supersaturated and precipitation occurs. If Qsp<KspQ_{sp} < K_{sp}, the solution is unsaturated. If Qsp=KspQ_{sp} = K_{sp}, the solution is saturated.

📐Formulae

Ksp=[Ay+]x[Bx−]yK_{sp} = [A^{y+}]^x [B^{x-}]^y

For ABexttypesalts(e.g.,AgClext):Ksp=s2\text{For } AB ext{ type salts (e.g., } AgCl ext{): } K_{sp} = s^2

For AB2extorA2Bexttypesalts(e.g.,PbCl2,Ag2CrO4ext):Ksp=4s3\text{For } AB_2 ext{ or } A_2B ext{ type salts (e.g., } PbCl_2, Ag_2CrO_4 ext{): } K_{sp} = 4s^3

For A2B3exttypesalts(e.g.,As2S3ext):Ksp=108s5\text{For } A_2B_3 ext{ type salts (e.g., } As_2S_3 ext{): } K_{sp} = 108s^5

General Relation: Ksp=(xs)x(ys)y=xxyys(x+y)\text{General Relation: } K_{sp} = (xs)^x (ys)^y = x^x y^y s^{(x+y)}

💡Examples

Problem 1:

The solubility of Ag2CrO4Ag_2CrO_4 is 1.3imes10−4 mol L−11.3 imes 10^{-4} \text{ mol L}^{-1} at 298 K298 \text{ K}. Calculate its solubility product.

Solution:

The dissociation of Ag2CrO4Ag_2CrO_4 is given by: Ag2CrO4(s)ightleftharpoons2Ag+(aq)+CrO42−(aq)Ag_2CrO_4 (s) ightleftharpoons 2Ag^+ (aq) + CrO_4^{2-} (aq). Given solubility s=1.3imes10−4 mol L−1s = 1.3 imes 10^{-4} \text{ mol L}^{-1}. From the stoichiometry, [Ag+]=2s[Ag^+] = 2s and [CrO42−]=s[CrO_4^{2-}] = s. Therefore, Ksp=[Ag+]2[CrO42−]=(2s)2(s)=4s3K_{sp} = [Ag^+]^2 [CrO_4^{2-}] = (2s)^2 (s) = 4s^3. Substituting ss: Ksp=4×(1.3imes10−4)3=4×2.197imes10−12=8.788imes10−12K_{sp} = 4 \times (1.3 imes 10^{-4})^3 = 4 \times 2.197 imes 10^{-12} = 8.788 imes 10^{-12}.

Explanation:

Since the salt is of the type A2BA_2B, the relationship Ksp=4s3K_{sp} = 4s^3 is applied directly after identifying the molar solubility.

Problem 2:

The KspK_{sp} of BaSO4BaSO_4 is 1.1imes10−101.1 imes 10^{-10}. Will a precipitate form when equal volumes of 0.0002 M BaCl20.0002 \text{ M } BaCl_2 and 0.0002 M Na2SO40.0002 \text{ M } Na_2SO_4 are mixed?

Solution:

When equal volumes are mixed, the concentration of each ion is halved. [Ba2+]=0.00022=1imes10−4 M[Ba^{2+}] = \frac{0.0002}{2} = 1 imes 10^{-4} \text{ M} and [SO42−]=0.00022=1imes10−4 M[SO_4^{2-}] = \frac{0.0002}{2} = 1 imes 10^{-4} \text{ M}. The Ionic Product Qsp=[Ba2+][SO42−]=(1imes10−4)×(1imes10−4)=1imes10−8Q_{sp} = [Ba^{2+}][SO_4^{2-}] = (1 imes 10^{-4}) \times (1 imes 10^{-4}) = 1 imes 10^{-8}. Since Qsp(1imes10−8)>Ksp(1.1imes10−10)Q_{sp} (1 imes 10^{-8}) > K_{sp} (1.1 imes 10^{-10}), a precipitate will form.

Explanation:

Precipitation is predicted by comparing the reaction quotient (ionic product) with the solubility product constant. Because the ionic product exceeds KspK_{sp}, the system shifts to form solid BaSO4BaSO_4.

Solubility Product Constant - Revision Notes & Key Formulas | CBSE Class 11 Chemistry