JEE Main 2022 Chemistry question
A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of [CH_{3}CH_{2}COO^{-}]/[CH_{3}CH_{2}COOH] required to make buffer is . Given : Ka(CH_{3}CH_{2}COOH) = 1.3 \times 10^{-5}
Answer: (2)
Solution
pH = + log([salt]/[acid]) gives 4 = −log(1.3 \times 10^{-5}) + \log ([CH_{3}CH_{2}COO^{-}]/[CH_{3}CH_{2}COOH]). So $ = 1.3 ×
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