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JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)

\(20\,mL\) of \(0.1\,M\,NaOH\) is added to \(50\,mL\) of \(0.1\,M\) acetic acid solution. The \(pH\) of the resulting solution is \(.......\times 10^{-2}\) (Nearest integer) \(\text { Given : pKa }\left( CH _3 COOH \right)=4.76\) \(\log 2=0.30\) \(\log 3=0.48\)

  1. A \(487\)
  2. B \(430\)
  3. C \(438\)
  4. D \(458\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(458\)

Step-by-step Solution

Detailed explanation

\(CH _3 COOH + NaOH \rightarrow CH _3 COONa + H _2 O\) \(pH = pKa +\log _{10} \frac{[ salt ]}{[\text { acid }]}\) \(pH =4.76+\log _{10} \frac{2}{3}\) \(pH =4.58=458 \times 10^{-2}\)
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