In an acidic buffer,0.25 M CH_3COOH is added | vedclass

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Question

(A) Given: $[Acid] = 0.25 \ M$,$[Salt] = 0.125 \ M$,$K_a = 1.8 	imes 10^{-5}$.First,calculate $pK_a$: $pK_a = -\log(K_a) = -\log(1.8 	imes 10^{-5})

Vedclass · Accepted Answer

$4.44$

Vedclass · Answer

(A) Given: $[Acid] = 0.25 \ M$,$[Salt] = 0.125 \ M$,$K_a = 1.8 	imes 10^{-5}$.First,calculate $pK_a$: $pK_a = -\log(K_a) = -\log(1.8 	imes 10^{-5}) = 5 - \log(1.8) = 5 - 0.255 = 4.745$.Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.$pH = 4.745 + \log \frac{0.125}{0.25} = 4.745 + \log(0.5)$.Since $\log(0.5) = -0.301$,$pH = 4.745 - 0.301 = 4.444$.Rounding to two decimal places,the $pH$ is $4.44$.

In an acidic buffer,$0.25 \ M$ $CH_3COOH$ is added to $0.125 \ M$ $NaCH_3COO$ solution. Calculate the $pH$ of the buffer. Given $K_a$ of $CH_3COOH = 1.8 \times 10^{-5}$. [Formula: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$]

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