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Equilibrium appeared 36 times across 3 years — 4.2% of Chemistry. This question is from Buffer Solutions.

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Consider a weak base 'B' of pKb = 5.699. 'x' mL of 0.02~M HCl and 'y' mL of 0.02~M weak base 'B' are mixed to make 100~mL of a buffer of pH 9 at 25°C. The values of 'x' and 'y' respectively are: [Given: 2 = 0.3010, 3 = 0.4771, 5 = 0.699]

Solution & Explanation

Related Formula
pOH = pKb + [ SaltBase]
Step 1: Find Active Moles

Reaction: B + HCl arrow BH^+ + Cl^- Moles of HCl = 0.02x mmol Moles of Base B = 0.02y mmol Since a buffer is formed, Base B is in excess. Moles of Salt formed (BH^+) = 0.02x Moles of Base remaining = 0.02y - 0.02x

Step 2: Apply Henderson-Hasselbalch Equation

Given pH = 9, so pOH = 14 - 9 = 5.

5 = 5.699 + [(0.02x)/(0.02y - 0.02x)] -0.699 = ((x)/(y - x))

We know 5 = 0.699, therefore (1/5) = -0.699.

(x)/(y - x) = (1)/(5) 5x = y - x y = 6x
Step 3: Solve for x and y

Total volume x + y = 100~mL.

x + 6x = 100 7x = 100 x = (100)/(7) ≈ 14.3~mL y = 100 - 14.3 = 85.7~mL
Pattern Recognition

Buffer mixing implies partial neutralization. The ratio of acid volume to base volume defines the logarithmic salt-to-base ratio. Using inverse logs directly maps variables to total volume restrictions.

Chapter Mix

Class 11 Chemistry: Equilibrium

Reference Study Guides

Equilibrium

More Equilibrium Previous-Year Questions — Page 8

Q62 jee_main_2024_31_jan_morning Equilibrium Constant
For the given reaction, choose the correct expression of KC from the following :- Fe(aq)³⁺ + SCN(aq)⁻ leftharpoons (FeSCN)(aq)²⁺
  • A. KC = [FeSCN²⁺][Fe³⁺][SCN⁻]
  • B. KC = [Fe³⁺][SCN⁻][FeSCN²⁺]
  • C. KC = [FeSCN²⁺][Fe³⁺]²[SCN⁻]²
  • D. KC = [FeSCN²⁺]²[Fe³⁺][SCN⁻]

Solution

Related Formula
KC = [Products][Reactants]
Core Logic
KC = Products ion conc.Reactants ion conc. KC = [FeSCN²⁺][Fe³⁺][SCN⁻]
Chapter Mix

Class 11 Chemistry: Equilibrium

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