UPSC Combined Geo-Scientist, Mains · 2026 · Mains Paper II
Q13(b). Buffer capacity of an acetic acid–acetate buffer
Question
Define buffer capacity (β) of a buffer solution. Show that β of an acetic acid-acetate buffer solution is maximum when [salt]/[acid] is one.
Answer in brief
β=(dC_b)/(d pH)=2.303 (C Kₐ[H₃O⁺])/((Kₐ+[H₃O⁺])²), maximum when [H₃O⁺]=Kₐ, i.e., [salt]/[acid] =1 and pH=pKₐ; then βₘₐₓ=0.576 C.
Key points
- Let C=[HA]+[A⁻] (total) and [A⁻]=C_b+[H₃O⁺]-[OH⁻]≈ C_b (buffer region). With Kₐ=[H₃O⁺][A⁻][HA]: [A⁻]=(CKₐ)/(Kₐ+[H₃O⁺]).
- β=d[A⁻]dpH=-2.303[H₃O⁺]d[A⁻]d[H₃O⁺] (since dpH=-(d[H₃O⁺])/(2.303[H₃O⁺])). Differentiating, d[A⁻]d[H₃O⁺]=-(CKₐ)/((Kₐ+[H₃O⁺])²) ⇒ β=2.303 (C Kₐ[H₃O⁺])/((Kₐ+[H₃O⁺])²)
- put x=[H₃O⁺] and set (dβ)/(dx)=0: d/dx[(x)/((Kₐ+x)²)]=((Kₐ+x)-2x)/((Kₐ+x)³)=0⇒ x=Kₐ.
- pH=pKₐ and [A⁻]=[HA]=C/2, so ([salt])/([acid])=1.
The full model answer
Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.
Definition: β is the number of moles of strong base (or acid) per litre needed to change the pH of the buffer by one unit: β=(dC_b)/(dpH) (strong base added, C_b).
- Let C=[HA]+[A⁻] (total) and [A⁻]=C_b+[H₃O⁺]-[OH⁻]≈ C_b (buffer region). With Kₐ=[H₃O⁺][A⁻][HA]: [A⁻]=(CKₐ)/(Kₐ+[H₃O⁺]).
- β=d[A⁻]dpH=-2.303[H₃O⁺]d[A⁻]d[H₃O⁺] (since dpH=-(d[H₃O⁺])/(2.303[H₃O⁺])). Differentiating, d[A⁻]d[H₃O⁺]=-(CKₐ)/((Kₐ+[H₃O⁺])²) ⇒ β=2.303 (C Kₐ[H₃O⁺])/((Kₐ+[H₃O⁺])²)
- Maximum: put x=[H₃O⁺] and set (dβ)/(dx)=0: d/dx[(x)/((Kₐ+x)²)]=((Kₐ+x)-2x)/((Kₐ+x)³)=0⇒ x=Kₐ.
- At [H₃O⁺]=Kₐ: pH=pKₐ and [A⁻]=[HA]=C/2, so ([salt])/([acid])=1.
- βₘₐₓ=2.303 (CKₐ²)/(4Kₐ²)=(2.303)/(4)C=0.576 C.
a buffer works best within pH=pKₐ±1 (ratio 10 to 0.1), where β is above one-third of its maximum; for acetic acid (pKₐ 4.76) the range is pH 3.8 to 5.8.
Coversbuffer capacityBuffer capacity of an acetic acid–acetate buffer
All parts of Question 13
- 13(a) ΔᵣG, ΔᵣH and ΔᵣS of the cell reaction10 marks
- 13(b) Buffer capacity of an acetic acid–acetate buffer10 marks