Question

20 mL of 0.1 M NaOH is added to 50 mL of 0.1 M acetic acid solution. The pH of the resulting solution is____ x 10^-2 (Nearest integer) 
Given: pKa (CH₃COOH) = 4.76 
log 2 = 0.30 
log 3 = 0.48

Answer 1

Correct Answer: 458

1. Reaction between CH\(_3\)COOH and NaOH: \[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}. \] Initially: \[ \text{Moles of CH}_3\text{COOH} = 50 \, \text{mL} \times 0.1 \, \text{M} = 5 \, \text{mmol}. \] \[ \text{Moles of NaOH} = 20 \, \text{mL} \times 0.1 \, \text{M} = 2 \, \text{mmol}. \] After reaction: - CH\(_3\)COOH remaining = \(5 - 2 = 3 \, \text{mmol}\). - CH\(_3\)COONa formed = \(2 \, \text{mmol}\). The resulting solution is a buffer solution containing CH\(_3\)COOH (acid) and CH\(_3\)COONa (salt). 2. Henderson-Hasselbalch equation: \[ \text{pH} = \text{p}K_a + \log_{10} \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right). \] Substitute the values: \[ \text{pH} = 4.76 + \log_{10} \left( \frac{2}{3} \right). \] 3. Simplify using logarithm values: \[ \log_{10} \left( \frac{2}{3} \right) = \log_{10}(2) - \log_{10}(3). \] Given \(\log 2 = 0.30\) and \(\log 3 = 0.48\): \[ \log_{10} \left( \frac{2}{3} \right) = 0.30 - 0.48 = -0.18. \] Substituting back: \[ \text{pH} = 4.76 - 0.18 = 4.58. \] 4. Convert to scientific notation: \[ \text{pH} = 4.58 \times 10^{-2}. \] 5. Final Answer: \[ \boxed{458 \times 10^{-2}} \]