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) \(\text { Given : pKa }\left( CH _3 COOH \right)=4.76\) \(\log 2=0.30\) \(\log 3=0.48\)
- A \(487\)
- B \(430\)
- C \(438\)
- D \(458\)
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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