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Equilibrium - Buffer Solutions

Grade 11CBSEChemistry

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

🔑Concepts

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A Buffer Solution is defined as a solution which resists a change in its pHpH value upon the addition of a small amount of strong acid or strong base, or upon dilution.

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Acidic Buffer: It is a mixture of a weak acid and its salt with a strong base (e.g., CH3COOHCH_3COOH and CH3COONaCH_3COONa). It maintains a pHpH in the acidic range (pH<7pH < 7).

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Basic Buffer: It is a mixture of a weak base and its salt with a strong acid (e.g., NH4OHNH_4OH and NH4ClNH_4Cl). It maintains a pHpH in the alkaline range (pH>7pH > 7).

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Buffer Action: The ability of the buffer solution to resist changes in pHpH is called buffer action. For an acidic buffer (HA/A−HA/A^-), added H+H^+ ions are consumed by A−A^-, and added OH−OH^- ions are neutralized by HAHA.

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Buffer Capacity: It is a measure of the effectiveness of a buffer in resisting pHpH changes. It is maximum when the concentration of the salt is equal to the concentration of the acid or base ([Salt]=[Acid][Salt] = [Acid] or [Salt]=[Base][Salt] = [Base]).

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The pHpH of a buffer solution is independent of the volume of the solution but depends on the ratio of the concentrations of the salt and the acid/base.

📐Formulae

pH=pKa+log⁡10[Salt][Acid]pH = pK_a + \log_{10} \frac{[Salt]}{[Acid]}

pOH=pKb+log⁡10[Salt][Base]pOH = pK_b + \log_{10} \frac{[Salt]}{[Base]}

pKa=−log⁡10KapK_a = -\log_{10} K_a

pKb=−log⁡10KbpK_b = -\log_{10} K_b

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

💡Examples

Problem 1:

Calculate the pHpH of a buffer solution containing 0.2 M0.2 \text{ M} acetic acid (CH3COOHCH_3COOH) and 0.5 M0.5 \text{ M} sodium acetate (CH3COONaCH_3COONa). The KaK_a for acetic acid is 1.8×10−51.8 \times 10^{-5}.

Solution:

  1. Find pKapK_a: pKa=−log⁡(1.8×10−5)=4.74pK_a = -\log(1.8 \times 10^{-5}) = 4.74.
  2. Use the Henderson-Hasselbalch equation: pH=pKa+log⁡[Salt][Acid]pH = pK_a + \log \frac{[Salt]}{[Acid]}.
  3. Substitute values: pH=4.74+log⁡0.50.2=4.74+log⁡(2.5)pH = 4.74 + \log \frac{0.5}{0.2} = 4.74 + \log(2.5).
  4. Since log⁡(2.5)≈0.398\log(2.5) \approx 0.398, pH=4.74+0.398=5.138pH = 4.74 + 0.398 = 5.138.

Explanation:

The pHpH is higher than the pKapK_a because the concentration of the salt (basic component) is higher than that of the acid.

Problem 2:

A basic buffer is prepared by mixing 0.1 M0.1 \text{ M} NH4OHNH_4OH and 0.1 M0.1 \text{ M} NH4ClNH_4Cl. If KbK_b for NH4OHNH_4OH is 1.8×10−51.8 \times 10^{-5}, what is the pHpH of the solution?

Solution:

  1. Find pKbpK_b: pKb=−log⁡(1.8×10−5)=4.74pK_b = -\log(1.8 \times 10^{-5}) = 4.74.
  2. Calculate pOHpOH: pOH=pKb+log⁡[NH4Cl][NH4OH]=4.74+log⁡0.10.1=4.74+log⁡(1)=4.74+0=4.74pOH = pK_b + \log \frac{[NH_4Cl]}{[NH_4OH]} = 4.74 + \log \frac{0.1}{0.1} = 4.74 + \log(1) = 4.74 + 0 = 4.74.
  3. Calculate pHpH: pH=14−pOH=14−4.74=9.26pH = 14 - pOH = 14 - 4.74 = 9.26.

Explanation:

When the concentrations of the weak base and its salt are equal, the pOHpOH of the buffer is equal to the pKbpK_b of the base.

Buffer Solutions - Revision Notes & Key Formulas | CBSE Class 11 Chemistry