A student needs to prepare a buffer solution | vedclass

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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: $K_{a}(CH_{3}CH_{2}COOH) = 1.3 \times 10^{-5}$

Similar Questions

$20 \ mL$ of $0.1 \ M \ H_2SO_4$ is added to $30 \ mL$ of $0.2 \ M \ NH_4OH$ solution. The $pH$ of the resultant mixture is [$pK_b$ of $NH_4OH = 4.7$].

Assertion : Mixture of $CH_3COOH$ and $CH_3COONH_4$ is an example of acidic buffer.
Reason : Acidic buffer contains equimolar mixture of a weak acid and its salt with weak base.

The dissociation constant of a weak acid is $1 \times 10^{-4}.$ In order to prepare a buffer solution with a $pH = 5,$ the $[Salt]/[Acid]$ ratio should be

Which of the following does not act as a buffer solution?

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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: $K_{a}(CH_{3}CH_{2}COOH) = 1.3 \times 10^{-5}$