Question

pH of a buffer solution decreases by \(0.02\) units when \(0.12\,g\) of acetic acid is added to \(250\,mL\) of a buffer solution of acetic acid and potassium acetate at \(27^\circ C\). The buffer capacity of the solution is

• A. \(0.1\)
• B. \(1\)
• C. \(10\)
• D. \(0.4\)

Hint:

Buffer capacity: \(\beta=\frac{\text{moles added}}{\Delta pH \times \text{volume (L)}}\). Convert mass into moles first.

Answer 1

The Correct Option is D

Step 1: Use definition of buffer capacity.
Buffer capacity \(\beta\) is:
\[ \beta = \frac{\Delta n}{\Delta pH \times V} \]
where \(\Delta n\) = moles of acid/base added, \(V\) = volume in litres.
Step 2: Calculate moles of acetic acid added.
Molar mass of \(CH_3COOH = 60\,g/mol\).
\[ \Delta n = \frac{0.12}{60} = 0.002\,mol \]
Step 3: Given pH change.
\[ \Delta pH = 0.02 \]
Volume:
\[ V = 250mL = 0.25L \]
Step 4: Compute buffer capacity.
\[ \beta = \frac{0.002}{0.02 \times 0.25} = \frac{0.002}{0.005} = 0.4 \]
Final Answer:
\[ \boxed{0.4} \]