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Equilibrium - Solubility Product and pH Scale

Grade 11ICSEChemistry

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

🔑Concepts

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The pHpH scale is a logarithmic scale used to specify the acidity or basicity of an aqueous solution. It is defined as pH=−log⁡10[H+]pH = -\log_{10}[H^+].

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The ionic product of water, denoted as KwK_w, is the product of the molar concentrations of hydrogen and hydroxyl ions. At 298 K298\text{ K}, Kw=[H+][OH−]=1.0×10−14K_w = [H^+][OH^-] = 1.0 \times 10^{-14}.

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For any aqueous solution at 298 K298\text{ K}, the sum of pHpH and pOHpOH is always equal to 1414, i.e., pH+pOH=14pH + pOH = 14.

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Solubility Product (KspK_{sp}) applies to sparingly soluble salts. It is the product of the molar concentrations of its ions in a saturated solution, each raised to the power of their stoichiometric coefficients.

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A salt AxByA_xB_y dissociates as AxBy(s)⇌xAy+(aq)+yBx−(aq)A_xB_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq). The expression is Ksp=[Ay+]x[Bx−]yK_{sp} = [A^{y+}]^x [B^{x-}]^y.

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The Ionic Product (QspQ_{sp}) is the product of ion concentrations at any state. Precipitation occurs only if Qsp>KspQ_{sp} > K_{sp}. If Qsp<KspQ_{sp} < K_{sp}, the solution is unsaturated.

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The Common Ion Effect states that the solubility of a sparingly soluble salt is suppressed by the addition of a strong electrolyte containing a common ion, shifting the equilibrium to the left according to Le Chatelier's Principle.

📐Formulae

pH=−log⁡10[H+]pH = -\log_{10}[H^+]

pOH=−log⁡10[OH−]pOH = -\log_{10}[OH^-]

Kw=[H+][OH−]=10−14 (at 25∘C)K_w = [H^+][OH^-] = 10^{-14} \text{ (at } 25^\circ\text{C)}

pH+pOH=pKw=14pH + pOH = pK_w = 14

Ksp=(xS)x⋅(yS)y=xxyyS(x+y)K_{sp} = (xS)^x \cdot (yS)^y = x^x y^y S^{(x+y)}

For AB type salt: Ksp=S2\text{For } AB \text{ type salt: } K_{sp} = S^2

For AB2 or A2B type salt: Ksp=4S3\text{For } AB_2 \text{ or } A_2B \text{ type salt: } K_{sp} = 4S^3

For A2B3 type salt: Ksp=108S5\text{For } A_2B_3 \text{ type salt: } K_{sp} = 108S^5

💡Examples

Problem 1:

Calculate the pHpH of a 0.001 M0.001\text{ M} solution of HClHCl.

Solution:

HClHCl is a strong acid, so [H+]=[HCl]=0.001 M=10−3 M[H^+] = [HCl] = 0.001\text{ M} = 10^{-3}\text{ M}. Using the formula pH=−log⁡[H+]pH = -\log[H^+], we get pH=−log⁡(10−3)=3pH = -\log(10^{-3}) = 3.

Explanation:

Since HClHCl dissociates completely, the concentration of H+H^+ ions is equal to the molarity of the acid. The log base 10 of 10−310^{-3} is −3-3, and the negative sign makes the pHpH positive.

Problem 2:

The solubility of AgClAgCl in water is 1.06×10−5 mol L−11.06 \times 10^{-5}\text{ mol L}^{-1} at 298 K298\text{ K}. Calculate its solubility product (KspK_{sp}).

Solution:

The dissociation is AgCl(s)⇌Ag+(aq)+Cl−(aq)AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq). Here, S=1.06×10−5S = 1.06 \times 10^{-5}. The formula is Ksp=[Ag+][Cl−]=S×S=S2K_{sp} = [Ag^+][Cl^-] = S \times S = S^2. Therefore, Ksp=(1.06×10−5)2=1.12×10−10K_{sp} = (1.06 \times 10^{-5})^2 = 1.12 \times 10^{-10}.

Explanation:

For a 1:11:1 electrolyte like AgClAgCl, the solubility product is simply the square of the molar solubility.

Problem 3:

Calculate the solubility of CaF2CaF_2 in water if its KspK_{sp} is 3.4×10−113.4 \times 10^{-11}.

Solution:

CaF2CaF_2 dissociates as CaF2⇌Ca2++2F−CaF_2 \rightleftharpoons Ca^{2+} + 2F^-. If solubility is SS, then [Ca2+]=S[Ca^{2+}] = S and [F−]=2S[F^-] = 2S. Ksp=[Ca2+][F−]2=S(2S)2=4S3K_{sp} = [Ca^{2+}][F^-]^2 = S(2S)^2 = 4S^3. 3.4×10−11=4S3  ⟹  S3=8.5×10−12  ⟹  S=8.5×10−123≈2.04×10−4 mol L−13.4 \times 10^{-11} = 4S^3 \implies S^3 = 8.5 \times 10^{-12} \implies S = \sqrt[3]{8.5 \times 10^{-12}} \approx 2.04 \times 10^{-4}\text{ mol L}^{-1}.

Explanation:

For AB2AB_2 type salts, the concentration of the second ion is twice the solubility, and this concentration is squared in the KspK_{sp} expression, leading to the 4S34S^3 relationship.

Solubility Product and pH Scale - Revision Notes & Key Formulas | ICSE Class 11 Chemistry