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(B) The Henderson-Hasselbalch equation for an acidic buffer is given by:$pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$First,calculate $pK_{a}$:$pK_{a} = -

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$0.13$

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(B) The Henderson-Hasselbalch equation for an acidic buffer is given by:$pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$First,calculate $pK_{a}$:$pK_{a} = -\log(K_{a}) = -\log(1.3 	imes 10^{-5}) = 5 - \log(1.3) \approx 5 - 0.1139 = 4.8861$Substitute the values into the equation:$4 = 4.8861 + \log \frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]}$$\log \frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]} = 4 - 4.8861 = -0.8861$Taking the antilog:$\frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]} = 10^{-0.8861} \approx 0.13$

$A$ student needs to prepare a buffer solution of propanoic acid and its sodium salt with $pH$ $4$. The ratio of $\frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]}$ required to make the buffer is .....
Given: $K_{a}(CH_{3}CH_{2}COOH) = 1.3 \times 10^{-5}$

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$A$ student needs to prepare a buffer solution of propanoic acid and its sodium salt with $pH$ $4$. The ratio of $\frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]}$ required to make the buffer is .....Given: $K_{a}(CH_{3}CH_{2}COOH) = 1.3 \times 10^{-5}$