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Equilibrium - Ionic Equilibrium (Acids, Bases, and Salts)

Grade 11ICSEChemistry

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

πŸ”‘Concepts

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Electrolytes: Substances that dissociate into ions in aqueous solution. Strong Electrolytes (e.g., HClHCl, NaOHNaOH, NaClNaCl) dissociate completely, while Weak Electrolytes (e.g., CH3COOHCH_3COOH, NH4OHNH_4OH) dissociate partially.

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Arrhenius Theory: Acids produce H+H^+ ions in water; Bases produce OHβˆ’OH^- ions in water.

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BrΓΈnsted-Lowry Theory: Acids are proton (H+H^+) donors; Bases are proton acceptors. This leads to the concept of Conjugate Acid-Base Pairs.

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Lewis Theory: Acids are electron pair acceptors; Bases are electron pair donors.

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Ostwald's Dilution Law: For a weak electrolyte with degree of dissociation Ξ±\alpha, K=CΞ±21βˆ’Ξ±K = \frac{C\alpha^2}{1-\alpha}. For very weak electrolytes, Ξ±=KC\alpha = \sqrt{\frac{K}{C}}.

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Ionic Product of Water (KwK_w): The product of molar concentrations of hydrogen and hydroxyl ions. At 298Β K298 \text{ K}, Kw=[H+][OHβˆ’]=1.0Γ—10βˆ’14Β mol2Lβˆ’2K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \text{ mol}^2\text{L}^{-2}.

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pH Scale: Defined as the negative logarithm of the hydronium ion concentration: pH=βˆ’log⁑10[H3O+]pH = -\log_{10}[H_3O^+].

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Common Ion Effect: The suppression of the degree of dissociation of a weak electrolyte by the addition of a strong electrolyte containing a common ion (e.g., adding CH3COONaCH_3COONa to CH3COOHCH_3COOH).

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Buffer Solutions: Solutions that resist changes in pHpH upon addition of small amounts of acid or base. Acidic buffer: Weak acid + its salt with a strong base. Basic buffer: Weak base + its salt with a strong acid.

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Solubility Product (KspK_{sp}): For a sparingly soluble salt AxByA_xB_y, the equilibrium is AxBy(s)β‡ŒxAy+(aq)+yBxβˆ’(aq)A_xB_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq). The Ksp=[Ay+]x[Bxβˆ’]yK_{sp} = [A^{y+}]^x [B^{x-}]^y at saturation.

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Salt Hydrolysis: The reaction of cations or anions of a salt with water to produce acidity or alkalinity. Salts of strong acids and strong bases do not undergo hydrolysis.

πŸ“Formulae

pH=βˆ’log⁑10[H+]pH = -\log_{10}[H^+]

pOH=βˆ’log⁑10[OHβˆ’]pOH = -\log_{10}[OH^-]

pH+pOH=pKw=14Β (atΒ 298Β K)pH + pOH = pK_w = 14 \text{ (at 298 K)}

Ka=CΞ±21βˆ’Ξ±β‰ˆCΞ±2β€…β€ŠβŸΉβ€…β€ŠΞ±=KaCK_a = \frac{C\alpha^2}{1-\alpha} \approx C\alpha^2 \implies \alpha = \sqrt{\frac{K_a}{C}}

[H+]=CΞ±=Kaβ‹…C[H^+] = C\alpha = \sqrt{K_a \cdot C}

pH=pKa+log⁑10[Salt][Acid] (Henderson-Hasselbalch for Acidic Buffer)pH = pK_a + \log_{10}\frac{[\text{Salt}]}{[\text{Acid}]} \text{ (Henderson-Hasselbalch for Acidic Buffer)}

pOH=pKb+log⁑10[Salt][Base] (Henderson-Hasselbalch for Basic Buffer)pOH = pK_b + \log_{10}\frac{[\text{Salt}]}{[\text{Base}]} \text{ (Henderson-Hasselbalch for Basic Buffer)}

Ksp=(xs)x(ys)y=xxyys(x+y)Β (whereΒ sΒ isΒ solubilityΒ inΒ mol/L)K_{sp} = (xs)^x (ys)^y = x^x y^y s^{(x+y)} \text{ (where } s \text{ is solubility in mol/L)}

Kh=KwKaΒ (HydrolysisΒ constantΒ forΒ saltΒ ofΒ WeakΒ AcidΒ andΒ StrongΒ Base)K_h = \frac{K_w}{K_a} \text{ (Hydrolysis constant for salt of Weak Acid and Strong Base)}

pH=7+12(pKa+log⁑C) (For salt of Weak Acid and Strong Base)pH = 7 + \frac{1}{2}(pK_a + \log C) \text{ (For salt of Weak Acid and Strong Base)}

πŸ’‘Examples

Problem 1:

Calculate the pHpH of a 0.1Β M0.1 \text{ M} solution of acetic acid (CH3COOHCH_3COOH) given that Ka=1.8Γ—10βˆ’5K_a = 1.8 \times 10^{-5}.

Solution:

  1. Use the formula [H+]=Kaβ‹…C[H^+] = \sqrt{K_a \cdot C}.
  2. [H+]=1.8Γ—10βˆ’5Γ—0.1=1.8Γ—10βˆ’6β‰ˆ1.34Γ—10βˆ’3Β M[H^+] = \sqrt{1.8 \times 10^{-5} \times 0.1} = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3} \text{ M}.
  3. pH=βˆ’log⁑(1.34Γ—10βˆ’3)=3βˆ’log⁑(1.34)β‰ˆ3βˆ’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, we use the simplified dissociation constant formula to find the hydrogen ion concentration, then convert it to pHpH using the logarithmic scale.

Problem 2:

The solubility of Ag2CrO4Ag_2CrO_4 is 1.3Γ—10βˆ’4Β molΒ Lβˆ’11.3 \times 10^{-4} \text{ mol L}^{-1}. Calculate its solubility product (KspK_{sp}).

Solution:

  1. Dissociation: Ag2CrO4(s)β‡Œ2Ag+(aq)+CrO42βˆ’(aq)Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq).
  2. If solubility is ss, then [Ag+]=2s[Ag^+] = 2s and [CrO42βˆ’]=s[CrO_4^{2-}] = s.
  3. Ksp=[Ag+][CrO42βˆ’]=(2s)2(s)=4s3K_{sp} = [Ag^+][CrO_4^{2-}] = (2s)^2(s) = 4s^3.
  4. Ksp=4Γ—(1.3Γ—10βˆ’4)3=4Γ—2.197Γ—10βˆ’12=8.788Γ—10βˆ’12K_{sp} = 4 \times (1.3 \times 10^{-4})^3 = 4 \times 2.197 \times 10^{-12} = 8.788 \times 10^{-12}.

Explanation:

The KspK_{sp} expression is derived from the stoichiometry of the salt's dissociation. For an A2BA_2B type salt, Ksp=4s3K_{sp} = 4s^3.

Ionic Equilibrium (Acids, Bases, and Salts) Revision - Class 11 Chemistry ICSE