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Equilibrium - Ionic Equilibrium in Solution

Grade 11CBSEChemistry

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

🔑Concepts

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Electrolytes are substances that conduct electricity in aqueous solution or molten state. Strong electrolytes like HClHCl or NaOHNaOH dissociate completely, while weak electrolytes like CH3COOHCH_3COOH or NH4OHNH_4OH exist in equilibrium with their ions.

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Arrhenius Theory defines an acid as a substance that increases [H+][H^+] in water and a base as one that increases [OH−][OH^-].

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Brønsted-Lowry Theory defines an acid as a proton (H+H^+) donor and a base as a proton acceptor. A conjugate acid-base pair differs only by a single proton, e.g., NH3NH_3 (base) and NH4+NH_4^+ (conjugate acid).

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Lewis Theory defines an acid as an electron pair acceptor (e.g., BF3BF_3, AlCl3AlCl_3) and a base as an electron pair donor (e.g., NH3NH_3, H2OH_2O).

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The Common Ion Effect is the suppression of the degree of dissociation of a weak electrolyte by the addition of a strong electrolyte containing a common ion, such as adding CH3COONaCH_3COONa to CH3COOHCH_3COOH.

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Buffer solutions resist changes in pHpH upon the addition of small amounts of acid or base. Acidic buffers consist of a weak acid and its salt with a strong base (e.g., CH3COOHCH_3COOH + CH3COONaCH_3COONa).

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Solubility Product (KspK_{sp}) is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution.

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Hydrolysis of salts involves the reaction of anions or cations of a salt with water to produce acidity or alkalinity. Salts of strong acids and strong bases (like NaClNaCl) do not undergo hydrolysis.

📐Formulae

pH=−log⁡10[H+]pH = -\log_{10}[H^+]

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

Kw=[H+][OH−]=1.0×10−14 at 298KK_w = [H^+][OH^-] = 1.0 \times 10^{-14} \text{ at } 298 K

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

Ka=cα21−α≈cα2 (for weak acids where α≪1)K_a = \frac{c \alpha^2}{1 - \alpha} \approx c \alpha^2 \text{ (for weak acids where } \alpha \ll 1)

α=Kac (Ostwald’s Dilution Law)\alpha = \sqrt{\frac{K_a}{c}} \text{ (Ostwald's Dilution Law)}

pH=pKa+log⁡[Salt][Acid] (Henderson-Hasselbalch Equation for Acidic Buffer)pH = pK_a + \log\frac{[\text{Salt}]}{[\text{Acid}]} \text{ (Henderson-Hasselbalch Equation for Acidic Buffer)}

Ksp=[Ay+]x[Bx−]y for a salt AxByK_{sp} = [A^{y+}]^x [B^{x-}]^y \text{ for a salt } A_xB_y

Kw=Ka×Kb (for a conjugate acid-base pair)K_w = K_a \times K_b \text{ (for a conjugate acid-base pair)}

💡Examples

Problem 1:

Calculate the pHpH of a 0.10M0.10 M solution of acetic acid (CH3COOHCH_3COOH), given that Ka=1.8×10−5K_a = 1.8 \times 10^{-5}.

Solution:

  1. Use the formula [H+]=Ka⋅c[H^+] = \sqrt{K_a \cdot c}.
  2. [H+]=(1.8×10−5)×0.10=1.8×10−6≈1.34×10−3M[H^+] = \sqrt{(1.8 \times 10^{-5}) \times 0.10} = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3} M.
  3. pH=−log⁡(1.34×10−3)=3−log⁡(1.34)≈3−0.127=2.873pH = -\log(1.34 \times 10^{-3}) = 3 - \log(1.34) \approx 3 - 0.127 = 2.873.

Explanation:

Since acetic acid is a weak acid, it only partially dissociates. We use the simplified equilibrium expression for [H+][H^+] and then apply the log function to find pHpH.

Problem 2:

The solubility of AgClAgCl (Ksp=1.6×10−10K_{sp} = 1.6 \times 10^{-10}) in water is to be determined. Calculate the molar solubility SS.

Solution:

  1. Write the dissociation equation: AgCl(s)⇌Ag+(aq)+Cl−(aq)AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq).
  2. Let SS be the solubility. Then [Ag+]=S[Ag^+] = S and [Cl−]=S[Cl^-] = S.
  3. Ksp=[Ag+][Cl−]=S2K_{sp} = [Ag^+][Cl^-] = S^2.
  4. S=Ksp=1.6×10−10=1.26×10−5 mol/LS = \sqrt{K_{sp}} = \sqrt{1.6 \times 10^{-10}} = 1.26 \times 10^{-5} \text{ mol/L}.

Explanation:

For a binary salt (ABAB type), the solubility product is the square of the molar solubility. Taking the square root of KspK_{sp} gives the concentration of ions in a saturated solution.

Ionic Equilibrium in Solution - Revision Notes & Key Formulas | CBSE Class 11 Chemistry