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Equilibrium - Ionization of Acids and Bases

Grade 11CBSEChemistry

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

πŸ”‘Concepts

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Arrhenius Concept: Acids are substances that dissociate in water to give H+H^+ ions, while bases provide OHβˆ’OH^- ions.

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BrΓΆnsted-Lowry Concept: An acid is a proton (H+H^+) donor, and a base is a proton acceptor. This leads to the concept of conjugate acid-base pairs, which differ only by one proton.

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Lewis Concept: Acids are electron-pair acceptors (e.g., BF3BF_3, AlCl3AlCl_3), and bases are electron-pair donors (e.g., NH3NH_3, H2OH_2O).

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

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The pHpH Scale: A logarithmic scale used to express the acidity or alkalinity of a solution. pH=βˆ’log⁑10[H+]pH = -\log_{10}[H^+].

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Ionization Constant (KaK_a and KbK_b): Measures the strength of weak acids and bases. For a weak acid HAHA, Ka=[H+][Aβˆ’][HA]K_a = \frac{[H^+][A^-]}{[HA]}.

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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.

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Buffer Solutions: Solutions that resist changes in pHpH on dilution or with the addition of small amounts of acid or alkali. Examples include mixtures of CH3COOHCH_3COOH and CH3COONaCH_3COONa.

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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 (pH=7pH = 7).

πŸ“Formulae

Kw=[H3O+][OHβˆ’]=10βˆ’14Β atΒ 298KK_w = [H_3O^+][OH^-] = 10^{-14} \text{ at } 298 K

pH=βˆ’log⁑10[H3O+]pH = -\log_{10}[H_3O^+]

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

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

Ka=CΞ±21βˆ’Ξ±β‰ˆCΞ±2Β (ifΒ Ξ±<0.05)K_a = \frac{C\alpha^2}{1-\alpha} \approx C\alpha^2 \text{ (if } \alpha < 0.05)

Ξ±=KaC\alpha = \sqrt{\frac{K_a}{C}}

pKa+pKb=pKw=14Β (forΒ conjugateΒ acid-baseΒ pair)pK_a + pK_b = pK_w = 14 \text{ (for conjugate acid-base pair)}

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

πŸ’‘Examples

Problem 1:

The ionization constant of HFHF is 3.2Γ—10βˆ’43.2 \times 10^{-4}. Calculate the degree of dissociation (Ξ±\alpha) and the pHpH of a 0.02M0.02 M solution of HFHF.

Solution:

Given Ka=3.2Γ—10βˆ’4K_a = 3.2 \times 10^{-4} and C=0.02MC = 0.02 M. Using Ξ±=KaC\alpha = \sqrt{\frac{K_a}{C}}, we get Ξ±=3.2Γ—10βˆ’40.02=0.016β‰ˆ0.126\alpha = \sqrt{\frac{3.2 \times 10^{-4}}{0.02}} = \sqrt{0.016} \approx 0.126. Now, [H+]=CΞ±=0.02Γ—0.126=2.52Γ—10βˆ’3M[H^+] = C\alpha = 0.02 \times 0.126 = 2.52 \times 10^{-3} M. pH=βˆ’log⁑(2.52Γ—10βˆ’3)=3βˆ’log⁑(2.52)β‰ˆ2.60pH = -\log(2.52 \times 10^{-3}) = 3 - \log(2.52) \approx 2.60.

Explanation:

Since KaK_a is relatively small, we use the approximation for Ξ±\alpha. The pHpH is then calculated from the concentration of H+H^+ ions produced by the partial ionization of the weak acid.

Problem 2:

Calculate the pHpH of a buffer solution containing 0.1M0.1 M CH3COOHCH_3COOH and 0.2M0.2 M CH3COONaCH_3COONa. (Given pKapK_a for CH3COOHCH_3COOH is 4.744.74)

Solution:

Using the Henderson-Hasselbalch equation: pH=pKa+log⁑[Salt][Acid]pH = pK_a + \log\frac{[\text{Salt}]}{[\text{Acid}]}. Substituting the values: pH=4.74+log⁑0.20.1=4.74+log⁑(2)=4.74+0.301=5.041pH = 4.74 + \log\frac{0.2}{0.1} = 4.74 + \log(2) = 4.74 + 0.301 = 5.041.

Explanation:

This example demonstrates how the pHpH of an acidic buffer is determined by the ratio of the salt and acid concentrations and the dissociation constant of the weak acid.

Ionization of Acids and Bases - Revision Notes & Key Formulas | CBSE Class 11 Chemistry