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Equilibrium - Ionization of Acids and Bases

Grade 11CBSEChemistry

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

🔑Concepts

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

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Brönsted-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 OH−OH^- ions in water. At 298K298 K, Kw=[H+][OH−]=1.0×10−14mol2L−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=−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+][A−][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).

📐Formulae

Kw=[H3O+][OH−]=10−14 at 298KK_w = [H_3O^+][OH^-] = 10^{-14} \text{ at } 298 K

pH=−log⁡10[H3O+]pH = -\log_{10}[H_3O^+]

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

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

Ka=Cα21−α≈Cα2 (if α<0.05)K_a = \frac{C\alpha^2}{1-\alpha} \approx C\alpha^2 \text{ (if } \alpha < 0.05)

α=KaC\alpha = \sqrt{\frac{K_a}{C}}

pKa+pKb=pKw=14 (for conjugate acid-base 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)}

💡Examples

Problem 1:

The ionization constant of HFHF is 3.2×10−43.2 \times 10^{-4}. Calculate the degree of dissociation (α\alpha) and the pHpH of a 0.02M0.02 M solution of HFHF.

Solution:

Given Ka=3.2×10−4K_a = 3.2 \times 10^{-4} and C=0.02MC = 0.02 M. Using α=KaC\alpha = \sqrt{\frac{K_a}{C}}, we get α=3.2×10−40.02=0.016≈0.126\alpha = \sqrt{\frac{3.2 \times 10^{-4}}{0.02}} = \sqrt{0.016} \approx 0.126. Now, [H+]=Cα=0.02×0.126=2.52×10−3M[H^+] = C\alpha = 0.02 \times 0.126 = 2.52 \times 10^{-3} M. pH=−log⁡(2.52×10−3)=3−log⁡(2.52)≈2.60pH = -\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 α\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.