Question

The pH of a mixture containing 100 mL of 0.5 M acetic acid and 50 mL of 0.2 M NaOH is (p\(K_a\) of CH\(_3\)COOH = 4.8).
(Given: \( \log 3 = 0.48 \), \( \log 4 = 0.60 \))

• A. 4.8
• B. 4.2
• C. 9.2
• D. 5.4

Hint:

For buffer mixtures, apply the Henderson-Hasselbalch equation: pH = pKa + log([Salt]/[Acid]).

Answer 1

The Correct Option is B

Step 1: Calculate moles of acetic acid and NaOH.
Acetic acid: \( 100 \text{ mL} \times 0.5 \text{ M} = 50 \text{ mmol} \)
NaOH: \( 50 \text{ mL} \times 0.2 \text{ M} = 10 \text{ mmol} \) Step 2: Weak acid-strong base reaction (partial neutralization)
\[ \text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \] After neutralization:
CH\(_3\)COOH remaining = \( 50 - 10 = 40 \) mmol
CH\(_3\)COO\(^-\) formed = \( 10 \) mmol Step 3: Use Henderson-Hasselbalch equation
\[ \text{pH} = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) = 4.8 + \log \left( \frac{10}{40} \right) = 4.8 + \log(0.25) \] \[ \log(0.25) = \log \left( \frac{1}{4} \right) = -\log(4) = -0.60 \Rightarrow \text{pH} = 4.8 - 0.60 = 4.2 \]